EK 1001 Physics Q 118

[Pisiform]

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anyone ??? i couldn't solve it

Thanks !

 

[Pisiform]

sweet, I got it, we needed to use a different formula which rarely shows up on the 'CAT

 

[TwoPaddles]

sweet, I got it, we needed to use a different formula which rarely shows up on the 'CAT

So the answer explanation is wrong ?

 

[Pisiform]

So the answer explanation is wrong ?

Actually, I couldn't understand the explanation at the back. As the projectile starts from a platform 15m high and then drops to the ground, I was having difficulty find the total time of flight (which was 3sec) which we could plug in the equation x = ut to find the range.

I went to my professor who gave me a new formula to find that out:

height = h(initial) + u(initial velocity)t + 1/2gt2

this gave me total time of flight through which I could find the range.

If anybody has some easier way, plz lemme know !

 

[Rabolisk]

Is it possible for you to post the problem and/or the relevant information?

 

[MCATMadness]

its a super basic problem actually...and the concept is tested quite a lot.

you are just using deltaY=Voy + 1/2at^2 to find time.
delta Y = 15; Voy = 0, etc

and plugging that into deltaX = VoxT to find the range.

no special formula needed...cant remember the exact problem but i remember it testing this concept in a basic way. if you want me to post the details i can tmw.

hope that helps

 

[Pisiform]

its a super basic problem actually...and the concept is tested quite a lot.

you are just using deltaY=Voy + 1/2at^2 to find time.
delta Y = 15; Voy = 0, etc

and plugging that into deltaX = VoxT to find the range.

no special formula needed...cant remember the exact problem but i remember it testing this concept in a basic way. if you want me to post the details i can tmw.

hope that helps

Can you please explain the above method in more detail. I really appreciate it. Here is the question:

Q - A projectile is launched at an angle of 30 degrees to the horizontal from a 15m platform. Its initial velocity is 20m/s. How far does it travel?

A - 52m

Thanks alot !

 

[Rabolisk]

You are asked find the range of the motion. The projectile stops moving horizontally when it hits the ground, y = 0. The vertical and horizontal components of the motion are related through time.

When dealing with vectors, it's important to note which direction is positive and which is negative. Naturally, I chose up as positive and right (the direction of horizontal motion of projectile) as positive.

y = y0 + v0yt + 1/2at^2
0 = 1/2at^2 + v0yt + y0 - y Rearrange
t = (-v0y - sqrt(voy^2-4(a/2)(y0-y)))/a Quadratic formula

We can discard the other root because physically it makes no sense.

Note: y0 = 15m y = 0m v0y = 20sin(30) m/s a = -g = -10

x = x0 + v0xt + 1/2at^2
x = v0xt Simplification
x = v0x((-v0y + sqrt(voy^2-4(a/2)(y0-y)))/a)
x = 20cos(30)((-20sin(30) - sqrt((20sin(30))^2-4(-5)(15)))/-10)
x = 51.96 m = 52 m

Note that I did not plug in values until the very end. Typically, it's a good method of solving physics problems, but for this problem, it may help to put numerical values in earlier (e.g. substitute 10 for 20sin(30)). The MCAT tends to have answers that work out well (like t = 3), so you won't get the rounding error often introduced when values are plugged in before the end.

 

[Chevy Chase Fan]

i thought you didn't get a calculator on the mcat? how are you suppose to do things with trig calculations? lol

 

[Pisiform]

You should memorize simple trig values like:

Sin 30
Cos 30
Sin 60
Cos 60
Sin 0
Cos 0
.... and so on and so forth

The number/digits on the MCAT are more freindly