Question

Q: Consider a person holding a ball with their right hand and they position their left hand 0.50 m away from their right. The person throws the ball from one hand to the other and the maximum height the ball reaches is 1 m above the hands. Find the magnitude and direction of the initial velocity of the ball.

A: V₀ = 4.46 m/s, 82.8° counterclockwise from the +x direction, OR 0.55 m/s i + 4.43 m/s j

so i have the answer, i just would like to know how to get there

Answer 1

For Vertical velocity

v² - u² =2gh

Here, v = 0 , h =1m

=> Uy² =2*9.8*1

=> Uy = 4.427 m/sec   -------->This is vertical component of velocity

Applying v = u +gt

=> t = Uy/g

          = 4.427/9.8

           = 0.451 sec   --------------> this is time taken by ball to reach maximum height

=> time taken by ball for whole journey = 0.451 *2

                                                                = 0.903 sec

For horizontal velocity Vx

Vx *0.903 = 0.5

=> Vx = 0.553 m/sec   -----------------> this is horizontal component of velocity

So, u*sin(theta) = 4.427

       u*cos(theta) = 0.553

=> tan(theta) = 8.005

=> theta = 82.87 degrees

=> u = 4.427/sin(82.87)

          = 4.46 m/sec

so, magnitude of initial velocity = 4.46 m/sec

Direction of initial velocity = 82.87° counterclockwise from the +x direction

=> Initial velocity = 0.553 m/s i + 4.427 m/s j