Question

A squirrel jumps into the air with a velocity of 4 m/s at an angle of 50%. What is the maximum height reached by the squirrel?

Answer 1

The maximum height reached by the squirrel : 0.479 m

Further explanation

Given

vo= 4 m/s

θ = 50 °

Required

The maximum height

Solution

Parabolic motion :

$\tt h_{max}=\dfrac{v_o^2sin^2\theta}{2.g}$

Input the value

$\tt h_{max}=\dfrac{4^2\times (sin~50)^2}{2\times 9.8}\\\\h_{max}=0.479~m$

or you can use

Find t from vt= vo sin θ - gt(negative sign=against gravity)⇒vt=0 at peak(the maximum height)

and input t to vertical component : y=voy.t-1/2gt²