Arobot on a basketball court needs to throw the ball in the hoop. the ball have a velocity of 10 m/s and an an angle of 60 degrees. the high of hoop is 0.75 meters and the horizontal distance between the robot and hoop is 5 meters. a) draw a digram.
b)what is the high that the robot needs to throw the ball from.

Initial velocity = 10 m/s

θ = 60°

This is the case of projectile motion

So the horizontal component of velocity 10 m/s  =  10 cosθ

u = 10 cosθ

u = 10 cos 60°

u=5 m/s

x= 5 m

So in the horizontal direction

x = u .t

5 = 5 .t

t = 1 sec  The vertical component of velocity 10 m/s = 10 sinθ

Vo= 10 sinθ

Vo= 10 sin 60°

Vo = 8.66 m/s

h=3.75 m

So height of robot = 3.75 - 0.75 m

   height of robot =3 m

prolly a beacuse the momentuem the car has increases while going down hill

i don't know that i agree with that but what's the question ?