A ball is thrown into the air with an upward velocity of 36 ft/s. Its height h in feet after t seconds is given by the function h=-16t^2+36t+5 In how many seconds does the ball reach its maximum height? Round to the nearest hundredth if necessary. What is the ball’s maximum height?
(1) Answers
you can use suvat equations to solve this s = height u = initial velocity = 36m/s v = final velocity = 0m/s a = acceleration = -9.81m/s^2 t = time taken v = u + at so (v-u)/a = t = -36/-9.81= 3.67 seconds (3sf) (v^2-u^2)/2a = 132.11m
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