Question

If an object is projected upward from ground level with an initial velocity of 144144 ft per​ sec, then its height in feet after t seconds is given by ​s(t)equals=minus−16tsquared2plus+144144t. Find the number of seconds it will take to reach its maximum height. What is this maximum​ height?

Answer 1

Answer:

4.5 s, 324 ft

Explanation:

The object is projected upward with an initial velocity of

$v_0 = 144 ft/s$

The equation that describes its height at time t is

$s(t) = -16t^2 + 144 t$ (1)

where t, the time, is measured in seconds.

In order to find the time it takes for the object to reach the maximum height, we must find an expression for its velocity at time t, which can be found by calculating the derivative of the position, s(t):

$v(t) = s'(t) = -32t +144$ (2)

At the maximum heigth, the vertical velocity is zero:

v(t) = 0

Substituting into the equation above, we find the corresponding time at which the object reaches the maximum height:

$0=-32t+144\\t=\frac{144}{32}=4.5 s$

And by substituting this value into eq.(1), we also find the maximum height:

$s(t) = -16(4.5)^2+144(4.5)=324 ft$