Julie throws a ball to her friend Sarah. The ball leaves Julie's hand a distance 1.5 meters above the ground with an initial speed of 16 m/s at an angle 47 degrees; with respect to the horizontal. Sarah catches the ball 1.5 meters above the ground.

What is the maximum height the ball goes above the ground?

I understand that the velocity should = 0, and I think I found the time to be 2.39 seconds, but when I plug everything in it always turns out wrong =(

I use the formula d=vt-1/2at^2 since v should = 0
I do d=-1/2at^2 but when I plug the values in the answer is wrong. Please help!

Question

Julie throws a ball to her friend Sarah. The ball leaves Julie's hand a distance 1.5 meters above the ground with an initial speed of 16 m/s at an angle 47 degrees; with respect to the horizontal. Sarah catches the ball 1.5 meters above the ground.

What is the maximum height the ball goes above the ground?

I  understand that the velocity should = 0, and I think I found the time to be 2.39 seconds, but when I plug everything in it always turns out wrong =(

I use the formula d=vt-1/2at^2 since v should = 0
I do d=-1/2at^2 but when I plug the values in the answer is wrong. Please help!

anonymous · Answer

Consider horizontal component:
Using $$v^2=u^2+2as$$$$0=(16\sin47^{0})^2+2(-9.81)s$$$$s=6.98m$$$$=7.0m(2s.f.)$$
Hence, maximum height is 8.5m