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
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
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