MCAT Physics

[ali1968]

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If an antelope is running at a speed of 10 mis, and can
maintain that horizontal velocity when it jumps, how high
must it jump in order to clear a horizqntal distance of 20
m..)

answer is 5 m
But how do you get it?

 

[krnpanda]

are you sure 5m is the correct answer?

Looks to me like it might not work out

if the person jumps 5m in the air, according to
x= 1/2 a t^2
5= 1/2 (10) t^2
10= 10 t^2
t=1

That person has 1 sec to clear 20m when he or she has a MAXIMUM horizontal speed 10 m/s. Maybe my logic is flawed somewhere?


Running at 10m/s horizontal, it needs a full 2 seconds to clear 20 m
x (vertical) = 1/2 a t^2
x= 1/2 (10) (2)^2
X= 1/2 (10) (4)
X= 20M

edit: on second thought, the calculations above might be for total distance traveled, so 20m would be going up and going down, so the maximum height might just be 10m

Then again, i suppose my math could be all wrong, but that's my guess

 

[Rabolisk]

are you sure 5m is the correct answer?

Looks to me like it might not work out

if the person jumps 5m in the air, according to
x= 1/2 a t^2
5= 1/2 (10) t^2
10= 10 t^2
t=1

That person has 1 sec to clear 20m when he or she has a MAXIMUM horizontal speed 10 m/s. Maybe my logic is flawed somewhere?


Running at 10m/s horizontal, it needs a full 2 seconds to clear 20 m
x (vertical) = 1/2 a t^2
x= 1/2 (10) (2)^2
X= 1/2 (10) (4)
X= 20M

edit: on second thought, the calculations above might be for total distance traveled, so 20m would be going up and going down, so the maximum height might just be 10m

Then again, i suppose my math could be all wrong, but that's my guess

I agree your math is wrong. Most importantly, you have to realize that acceleration is a vector.

The antelope has a constant horizontal velocity, and its range is given. That can be related to the the vertical component of motion. You're looking for vertical displacement in half of the total time. Relate the known variables (one of them is time) to the unknown variable, displacement.

 

[Pons Asinorum]

if the person jumps 5m in the air, according to
x= 1/2 a t^2
5= 1/2 (10) t^2
10= 10 t^2
t=1
Maybe my logic is flawed somewhere?

Yes, logic flaw. x in your calculations is 5, but there are two iterations. 5m going up, and 5m coming down. Total time will be 2 seconds. Apply that time to the constant horizontal velocity to figure out horizontal displacement.

 

[krnpanda]

Yes, logic flaw. x in your calculations is 5, but there are two iterations. 5m going up, and 5m coming down. Total time will be 2 seconds. Apply that time to the constant horizontal velocity to figure out horizontal displacement.

bloody hell, i stand corrected.

 

[seep]

You don't even need to remember any formulas for this, or even put pencil to paper.

All you have to remember is that horizontal and vertical displacement are independent. The path to the answer then is easy:

It takes 2 seconds for the deer to travel 20 horizontal meters. He is therefore rising for the first second and falling for the next second. At the start of his fall his vertical velocity was 0 m/s, and at the end of his fall it was 9.8 m/s. Therefore his average velocity while falling was 4.9 m/s. Since he fell for 1 second, his fall began at 4.9 meters above the ground.

(though deer contract their legs during flight, so he actually cleared almost 6 meters 😀)

 

[BerkReviewTeach]

If an antelope is running at a speed of 10 mis, and can maintain that horizontal velocity when it jumps, how high must it jump in order to clear a horizqntal distance of 20m..)

answer is 5 m
But how do you get it?

Sorry for the tooting of the horn here, but BR has got a great way to get this right in 5 seconds.

First off, we need to assume it is a smart antelope and knows to take off at a 45-degree angle. Page 30 in the physics book shows that for a 45-degree launch, the range is 4x the max height (by geometry). Page 30 shows the values for 30-degrees and 60-degrees too, but that's not important here. It's a nice range trick to know.

To solve it legitimately, we need krnpanda's math. Thankfully the question writers made the question easy by saying the antelope maintains its x-speed, so 20m at a rate of 10 m/s will take 2 s of flight time. As Pons pointed out, that's 1s of ascent and 1s of descent.

In 1 second, an object will fall 5m in free fall, so 5m must be the max height.

If you happen to have access to a BR physics book, check out page 21 for the free fall math tricks and page 30 for the range trick. The pictures really help.

 

[PB13]

d=v/t
find total time =2

divide by 1/2 for time going up or down

V=Vo + a t
V=10

h= V^2/2g
=5

 

[desperadoflakes]

Here's how I would do it quickly, with intuition, and without formulas:

You know horizontal velocity is 10 m/s, and you have to clear 20 m. From this, you should ask "What is the flight time needed?"
10 m/s * 20 m = 2 second flight time. That means the flight is divided into 1 second up, and 1 second down.
What max. height do you need for a 2 second flight? It is the same as the vertical distance you'll fall in a 1 second drop. We know that when a = g, you fall 5 meters in one second. Thus the answer is 5m.