translational motion question EK

[sheenafard]

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there is a question that states ,

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

The answer is 5 m and i arrived at that answer using the formula V = sqrt (2gh) and solved for h.

however the solution to the problem arrived at the answer (5m) using x=x0+v0t+1/2at^2 and v^2=v0^2+2ax

is the way i did it correct or did i arrive at the right answer coincidentally?

 

[icedragon]

From what you stated, I think it was a coincidence that you ended up with 5.

The way that I would approach this problem is two fold.

First I would determine how long it would take for the antelope to travel 20 meters, and since he's going 10 m/s, it would take 2 seconds.

So then the problem becomes one where you can only focus on the vertical, if you launch say a ball up, how long would it take for it to touch the ground again? Well it would take the same amount of time to travel up as it would to travel down again which is 1 second up and 1 second down.

Taking this knowledge, we can plug it into x = 1/2 a (t)^2 with t=1 second and a = 10 m/s^2 and we end up with 5 m.

 

[Jepstein30]

From what you stated, I think it was a coincidence that you ended up with 5.

The way that I would approach this problem is two fold.

First I would determine how long it would take for the antelope to travel 20 meters, and since he's going 10 m/s, it would take 2 seconds.

So then the problem becomes one where you can only focus on the vertical, if you launch say a ball up, how long would it take for it to touch the ground again? Well it would take the same amount of time to travel up as it would to travel down again which is 1 second up and 1 second down.

Taking this knowledge, we can plug it into x = 1/2 a (t)^2 with t=1 second and a = 10 m/s^2 and we end up with 5 m.

It was a coincidence.

Think of the same problem with different numbers: running 20 m/s to travel 20 m

Using icedragon's correct strategy,

t = .5 s (to peak)
y = .5(a)t^2 = .5*10*.5^2 = 1.25 m

using your strategy,

V = sqrt(2gh)
(V^2)/2g = h
h = (20^2)/(2*10) = 400/20 = 20 m

it just worked with those numbers