Projectile Motion at an Angle

Question

" 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?"

So far I tried to get initial velocity using the formula :

Vf^2 = Vi^2 + 2 (a) (d)

0 = Vi^2 + 2 (-9.8 m/s^2) (3.7 m)

Answer 1

The equation you've indicated above will give you v_i = 8.5 m/s, but this is too low by a factor of √2.

The correct speed is 12 m/s. The reason why is explained below.

You have the right idea, but the equation

(v_fy)² = 0 = (v_iy)² – 2gh

applies only to the vertical component of velocity (v_y).

So solving this gives you the vertical component of the initial velocity (v_iy)

v_iy = √(2gh)

The is related to the initial velocity the puma leaves the ground (v_i) by the formula

v_iy = v_i sin(θ)

When θ = 45°, sin(θ) = 1/√2 so

v_iy = v_i/√2 which means that the velocity you're looking for is

v_i = √2 × v_iy = (√2)(√(2gh)) = 2√(gh) = 2√(9.8 × 3.7) = 12 m/s