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?

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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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