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)=(v_o^2sin^2\theta)/(2.g)`

Input the value

`\tt h_(max)=(4^2* (sin~50)^2)/(2* 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