EK Physics Lecture 1 Q19

[BCTbro09]

After Lecture 1 projectile motion section, #20 seems to either be tricky or I'm just not seeing how they got the answer. Can anyone explain this to me the easiest way possible please? The question is...

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?

Thanks.

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[BCTbro09]

sorry, it's #19 after lecture 1.

 

[Lukkie]

the answer key does a good job breaking it down, what are you having a problem with specifically.

 

[engineeredout]

The antelop has to jump so that after one second its y velocity is 0.

Vf = vi + at

0 = vf -10*1 (acceleration is negative)

vi = 10


Vf^2 = vi^2 + 2aD

100 = 2*10*D (vf = 0, acceleration is negative)

D = 5m.


Dunno how to make it clearer than that.