Physics examkrackers 1001

[crazy person]

HI ALL !!! I have 3 questions on kinematics from examkrackers 1001 physics q's that I'm having problems getting the right answer. Can someone please explain to me how you are coming up with the right solutions? I post it the right answers at the bottom. Thanks !


105) An object dropped from a height of 13m strikes the ground at 16m/s. In order for the object to strike the ground at 32m/s, it must be dropped from:

a) 18m
b) 26m
c) 39m
d) 52m

I keep coming up with a total different number and I don't know if i'm plugging the right number for the right variable or am I supposed to be using a different equation. I'm using V^2 = V0^2 + 2ad

119) A projectile is launched from a 25m platform at an angle of 30 degrees to the horizontal. Its initial velocity is 40m/s. How long is it in the air?

a) 2s
b) 3S
c)4s
d)5s

for this problem I used d= vot +1/2at^2 but i ended up with a quadratic equation and I know that's not right. again am i using the wrong equation or the set up?

121) A projectile launched over level ground reaches its maximum height in 10 seconds. Approximately what was the range of the projectile if it was launched with a speed of 200m/s?

a) 1700m
b) 2000m
c) 3400m
d)4000m

for this problem i used the range equation X=VXT and got 2000m which is incorrect.


answers:

105)D
119)D
121)C

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[Temperature101]

HI ALL !!! I have 3 questions on kinematics from examkrackers 1001 physics q's that I'm having problems getting the right answer. Can someone please explain to me how you are coming up with the right solutions? I post it the right answers at the bottom. Thanks !


105) An object dropped from a height of 13m strikes the ground at 16m/s. In order for the object to strike the ground at 32m/s, it must be dropped from:

a) 18m
b) 26m
c) 39m
d) 52m

I keep coming up with a total different number and I don't know if i'm plugging the right number for the right variable or am I supposed to be using a different equation. I'm using V^2 = V0^2 + 2ad

119) A projectile is launched from a 25m platform at an angle of 30 degrees to the horizontal. Its initial velocity is 40m/s. How long is it in the air?

a) 2s
b) 3S
c)4s
d)5s

for this problem I used d= vot +1/2at^2 but i ended up with a quadratic equation and I know that's not right. again am i using the wrong equation or the set up?

121) A projectile launched over level ground reaches its maximum height in 10 seconds. Approximately what was the range of the projectile if it was launched with a speed of 200m/s?

a) 1700m
b) 2000m
c) 3400m
d)4000m

for this problem i used the range equation X=VXT and got 2000m which is incorrect.


answers:

105)D
119)D
121)C

Stop creating many threads asking for the same questions... I answered them in the other thread. If you don't understand my explanation, you can bump the thread and ask for another opinion. Here is again...


You can do a ratio for the first one... V^2/V*^2 = d/d* (accelerations is the same and Vo = 0 since the objects were dropped from rest. You solve for d* = (V*^2)x(d)/V^2 = 32^2 x 13/16^2 ------> d = 52m


For the second one, you might have to use the quadratic equation since the projectile was not lunch from the ground... I will let someone else take a crack at it...They might be able to do it without using the quadratic equation.

The third one, you have to find the angle the projectile was launched, which is unlikely to appear in the MCAT. In order to find the angle, you have to find initial vertical velocity...Vy = Voy - gt . When the projectile reach it max height, it experiences an acceleration due to gravity (10m/s^2) to come down and its Vy = 0 m/s... So Voy = gt = 10 x 10 = 100 m/s.. To find the angle sin(theta) = Voy/Vo , which is 30 degrees. Then use that angle to calculate Vox = cos (theta)xVo = (cos 30 degrees)( 200) = 0.86 x 200 = 172 m/s ........ Then X = Vox (20 minutes) since the projectile takes 10 minutes to reach max height, it will take another 10 minutes to come down. X ~3440 meters

 

[crazy person]

I didn't even noticed your explanations until now but I do understand your explanations; however, for # 105, examcrakers is using the equation v= sqrt(2gh) and i don't knw why they are even using this equation.

for 119, I guess I'll have to wait and see who can answer this equation without using the quadratic equation

for the last one #121, you didnt't even use the 100m/s for the y direction because we don't need it, right? you only used the 200m/s for the x directin since that's what we want to find the range, right?

thanks

 

[Temperature101]

I didn't even noticed your explanations until now but I do understand your explanations; however, for # 105, examcrakers is using the equation v= sqrt(2gh) and i don't knw why they are even using this equation.

for 119, I guess I'll have to wait and see who can answer this equation without using the quadratic equation

for the last one #121, you didnt't even use the 100m/s for the y direction because we don't need it, right? you only used the 200m/s for the x directin since that's what we want to find the range, right?

thanks

You only have to use Vox for the range.... I use Voy to find the the angle the projective was launched.

For #105, I also use V = sqrt(2gh) but using it in one step. Also, to make it simple, you can find the acceleration for the first object which I think is 9. 84 m/s2 and use that acceleration to find the distance for the second object.

 

[Temperature101]

Are you using only EK physics for content review? If so. I suggest to supplement it with NOVA physics because conceptually NOVA is one of the best outhere IMO.

 

[crazy person]

Thanks your explanations were very helpful.

I'm using it with TPR. I was thinking about buying TBR though. People have said they all scored 30's thanks to TBR so that's why I was thinking about switching. Do you think Nova is better than TBR?

 

[Temperature101]

Thanks your explanations were very helpful.

I'm using it with TPR. I was thinking about buying TBR though. People have said they all scored 30's thanks to TBR so that's why I was thinking about switching. Do you think Nova is better than TBR?

TBR is better in term of having so many practice problems but NOVA is good in term of explaining basic stuff. I think overall TBR is better because after you are done doing all these practice problems. you are bound to get the basics down IMO.

 

[mehc012]

HI ALL !!! I have 3 questions on kinematics from examkrackers 1001 physics q's that I'm having problems getting the right answer. Can someone please explain to me how you are coming up with the right solutions? I post it the right answers at the bottom. Thanks !


105) An object dropped from a height of 13m strikes the ground at 16m/s. In order for the object to strike the ground at 32m/s, it must be dropped from:

a) 18m
b) 26m
c) 39m
d) 52m

I keep coming up with a total different number and I don't know if i'm plugging the right number for the right variable or am I supposed to be using a different equation. I'm using V^2 = V0^2 + 2ad

119) A projectile is launched from a 25m platform at an angle of 30 degrees to the horizontal. Its initial velocity is 40m/s. How long is it in the air?

a) 2s
b) 3S
c)4s
d)5s

for this problem I used d= vot +1/2at^2 but i ended up with a quadratic equation and I know that's not right. again am i using the wrong equation or the set up?

121) A projectile launched over level ground reaches its maximum height in 10 seconds. Approximately what was the range of the projectile if it was launched with a speed of 200m/s?

a) 1700m
b) 2000m
c) 3400m
d)4000m

for this problem i used the range equation X=VXT and got 2000m which is incorrect.


answers:

105)D
119)D
121)C

I didn't even noticed your explanations until now but I do understand your explanations; however, for # 105, examcrakers is using the equation v= sqrt(2gh) and i don't knw why they are even using this equation.

for 119, I guess I'll have to wait and see who can answer this equation without using the quadratic equation

for the last one #121, you didnt't even use the 100m/s for the y direction because we don't need it, right? you only used the 200m/s for the x directin since that's what we want to find the range, right?

thanks

For 105, they're using energy conservation instead of Newton's laws...Ugrav is converted into KE, so
mgh = ½mv²
2gh = v²
v = √(2gh)
though really it'd be more useful to rearrange it to solve for h:
h = v²/2g

For 119, you can get away without using the quadratic because of the answer choices.
Step 1: find the initial vertical velocity
Vy = vsin30
Vy = ½v
Vy = 20m/s
Step 2: find how long it takes to reach the height it was launched from, knowing that the vertical velocity there will simply be the negative of the initial vertical velocity
Vf = Vo + at
-20m/s = 20m/s -10m/s² * t
-40m/s = -10m/s * t
t = 4s
Step 3: Remember that it will take some time to finish falling, aka total t > 4s
Answer must be d.

If you wanted a precise answer past that, you could use energy conservation to find the final velocity:
½mV1² + mg*25 = ½mV2²
(-20m/s)² + 2*10m/s² * 25m = V2²
400 m²/s² + 500m²/s² = V2²
900m²/s² = V2²
V2 = ±30m/s (we know it's negative, so -30m/s)
Then calculate the add'l time after the projectile reaches the height it was launched from until it hits the ground:
V2 = V1 - gt
-30m/s = -20m/s -10m/s * t
t = 1s
then add it to the 4s you found earlier
4s + 1s = 5s

 

[Postictal Raiden]

As the mehc012 showed, there are two ways to get the answer. One is by POE, my favorite method, and the second by determining the correct answer.

In POE, determine how long it will take the object to reach its maximum height and multiply that by 2 to get the the time duration for the entire trip, which is 4s. Then you know that it will take additional time to travel the the 25m from the platform level down to the ground level.

The long way goes like this:

first, find out how long it takes to get to maximum height.
second, find the vertical distance traveled (20m)
third, calculate how long it takes to drop form max height to ground level, 45 m drop.
fourth, add the time it takes to reach max height to the time it takes to drop from max height.

 

[Temperature101]

As the mehc012 showed, there are two ways to get the answer. One is by POE, my favorite method, and the second by determining the correct answer.

In POE, determine how long it will take the object to reach its maximum height and multiply that by 2 to get the the time duration for the entire trip, which is 4s. Then you know that it will take additional time to travel the the 25m from the platform level down to the ground level.

The long way goes like this:

first, find out how long it takes to get to maximum height.
second, find the vertical distance traveled (20m)
third, calculate how long it takes to drop form max height to ground level, 45 m drop.
fourth, add the time it takes to reach max height to the time it takes to drop from max height.

Very unlikely to get a problem that requires that many steps in the MCAT...However, If for some reason AAMC gives a problem like that, the quickest way to do it is to do what you said.

 

[mehc012]

Very unlikely to get a problem that requires that many steps in the MCAT...However, If for some reason AAMC gives a problem like that, the quickest way to do it is to do what you said.

That's why I greyed out the 'to get a more precise answer' portion! I also just found the time until it falls back to launch level, that way you don't have to remember to double it

 

[adrakdavra]

short cut save time
Tup = Vy / g = 20/2 = 2 sec up and more then 2 sec down , so yes D sound good

 

[mehc012]

short cut save time
Tup = Vy / g = 20/2 = 2 sec up and more then 2 sec down , so yes D sound good

That's the same exact equation that we were using.

 

[adrakdavra]

Sweetheart, all I am saying time is money, and you know money never lies

 

[mehc012]

Sweetheart, all I am saying time is money, and you know money never lies

Not disagreeing...just pointing out that you posted the EXACT same equation and explanation, you were just less detailed with your description. And while time is good, thorough explanations --> better understanding --> better and quicker application of the concept on the exam.

 

[dudewheresmymd]

Is there a typo? The answer to 119 is 4s, but d) says 5 s? It's 4 s (answer choice c) right?

 

[milski]

Is there a typo? The answer to 119 is 4s, but d) says 5 s? It's 4 s (answer choice c) right?

25 m ramp - goes up some (2s), falls down the same (2s) and then falls done 25 m more (>0s) -> the answer is anything larger than 4s, aka d

 

[dudewheresmymd]

Oops missed the 25 m platform part. Thanks.

 

[crazy person]

Thanks to everyone !! and thank you mehc 012 for the thorough explanations ! I didn't even think about the energy conservation at all. but now it makes more sense. thanks !

 

[shahcoco]

Stop creating many threads asking for the same questions... I answered them in the other thread. If you don't understand my explanation, you can bump the thread and ask for another opinion. Here is again...


You can do a ratio for the first one... V^2/V*^2 = d/d* (accelerations is the same and Vo = 0 since the objects were dropped from rest. You solve for d* = (V*^2)x(d)/V^2 = 32^2 x 13/16^2 ------> d = 52m


For the second one, you might have to use the quadratic equation since the projectile was not lunch from the ground... I will let someone else take a crack at it...They might be able to do it without using the quadratic equation.

The third one, you have to find the angle the projectile was launched, which is unlikely to appear in the MCAT. In order to find the angle, you have to find initial vertical velocity...Vy = Voy - gt . When the projectile reach it max height, it experiences an acceleration due to gravity (10m/s^2) to come down and its Vy = 0 m/s... So Voy = gt = 10 x 10 = 100 m/s.. To find the angle sin(theta) = Voy/Vo , which is 30 degrees. Then use that angle to calculate Vox = cos (theta)xVo = (cos 30 degrees)( 200) = 0.86 x 200 = 172 m/s ........ Then X = Vox (20 minutes) since the projectile takes 10 minutes to reach max height, it will take another 10 minutes to come down. X ~3440 meters

For #121, is there a short cut/POE method to finding that answer?

 

[Temperature101]

For #121, is there a short cut/POE method to finding that answer?

No short cut IMO... I don't know if one can use POE to get the answer... Don't worry about questions like that... EK physics can get crazy sometimes.