Answered

The best leaper in the animal kingdom is the puma, which can jump to a height of 3.7 m when leaving the ground at an angle of 45°. With what speed must the animal leave the ground to reach that height?

Answer :

Speed required is 12.05 m/s

Explanation:

Let the velocity of puma be a.

Consider the vertical motion of puma

We have equation of motion v² = u² + 2as

Initial velocity, u = asin45

Acceleration, a = -9.81 m/s²  

Final velocity, v = 0 m/s  

Displacement, s = 3.7 m

Substituting  

v² = u² + 2as

0² = (asin45)² + 2 x -9.81 x 3.7

a² = 145.19

a = 12.05 m/s

Speed required is 12.05 m/s

Other Questions