Question
A 0.175-lb ball is thrown upward with an initial velocity of 75.0 ft/s. What is the
maximum height reached by the ball?
Would I need to set it up like
√2*75.0*-32.2/0/175?
maximum height reached by the ball?
Would I need to set it up like
√2*75.0*-32.2/0/175?
Answers
Henry
V^2 = Vo^2 + 2g*h = 0 @ max ht.
h = -(Vo^2)/2g = -(75^2)/-64 = 87.9 Ft.
h = -(Vo^2)/2g = -(75^2)/-64 = 87.9 Ft.