A ball is thrown in the air from the top of a building Its h

A ball is thrown in the air from the top of a building. Its height, in meters above ground, as a function of time, in seconds, is given by h(t) = -4.9t^2 + 24t + 10. How long does it take to reach maximum height? () ______s

Solution

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Given height of ball at any time t after it is thrown in air h(t) = -4.9t^2 + 24t + 10

Now in order to calculate the time when the ball reaches maximum height we need to differentiate h(t) w.r.t. t

dh(t) / dt = -9.8t + 24 = 0

9.8t = 24

t = 2.44898 secs

Now in order to confirm that that the time calculates corresponds to maximum height we need to double differentiate h(t) and if the result is negative then it means calclated t corresponds to local maxima.

d2h/dt2 = -9.8

Hence t = 2.44898 corresponds to local maxima

Solution