Plz help Physic people?!?

Question

A skateboarder is slowing down at a rate of .70 m/s^2. At the moment he is moving forward at 1.5 m/s, he throws a basketball upward so that it reaches a height of 3.0 m, and then he catches it at the same level it was thrown without changing hie position on the skateboard. Determine the vertical and horizontal components of the ball's velocity relative to the skateboard when the ball left his hand.

Please help me with this one....I'm stumped! Thanks!

Answer 1

The distance traveled by the ball is the same as the distance traveled by the skateboarder. The lapsed time is the same too.
Major difference the skateboarder is traveling in horizontal direction with a negative acceleration while the ball is traveling it the horizontal direction with a constant velocity.

S=V0t'+ 0.5 at'^2
since
t= time to go up for the ball = time to come down=1/2 t'

t= sqrt(2 h/g)
S=2V0t + at^2
t= sqrt(2 x 3/ 9.81)=0.78 s
Vh= S/t'=S/2t= (2V0 t + at^2) /2t
Vh=V0+0.5 at
Vh= 1.5 -0.5 x 0.70 x .78=
Vh=1.23 m/s

Vv= gt
Vv= 9.81 x 0.78=7.7 m/s