SOLUTION: A ball is thrown upward with an initial velocity of 30 feet per second from a point that is 24 feet above the ground. The height (h) in feet of the ball at time t (in second) is gi
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Question 1166061: A ball is thrown upward with an initial velocity of 30 feet per second from a point that is 24 feet above the ground. The height (h) in feet of the ball at time t (in second) is given by the equation: h(t)= -16t^2+30t+24
a) at what time did the ball reach its maximum height?
b) what was the maximum height? Answer by htmentor(1343) (Show Source):
You can put this solution on YOUR website! The maximum height is reached when the derivative dh/dt = 0:
dh/dt = -32t + 30 = 0
This gives t = 30/32 = 0.9375 s
h(0.9375) = -16*(0.9375)^2 + 30*0.9375 + 24 = 38.063 ft
