Question

A ball is thrown directly upward from a height of 3 ft with an initial velocity of 28 ​ft/sec. the function ​s(t)equalsminus16tsquaredplus28tplus3 gives the height of the​ ball, in​ feet, t seconds after it has been thrown. determine the time at which the ball reaches its maximum height and find the maximum height.

Answer 1

Answer:

$t = 0.875 s$

$s_{max} = 15.25 ft$

Explanation:

Position of the ball as a function of time is given as

$s(t) = s_o + v_y t + \frac{1}{2}at^2$

$s(t) = 3 + 28 t - 16 t^2$

now we know that when ball will attain maximum height then the differentiation of the position with respect to time will become zero

so we have

$\frac{ds}{dt} = 0$

$0 = 0 + 28 - 32 t$

$t = 0.875 s$

now the maximum height is given as

$s_{max} = 3 + 28(0.875) - 16(0.875)^2$

$s_{max} = 15.25 ft$