A projectile is launched upward with a velocity of 288 feet per second from the top of a 65 foot structure . What is the maximum height attained by the projectile ?

[tex]d = vt + \frac{1}{2} g {t}^{2} \\ 0 = 288t - 5 {t}^{2} \\ t = 0 \\ or \\ t = \frac{ 288}{5} \\ \frac{t}{2} = \frac{ 288}{10} = 2.88 \: sec \\ d = 288(2.88) + 5 ({2.88}^{2} ) \\ d = 829.44 + 41.472 = 870.912 \\ h = 870.912 + 65 = 935.912 \: ft[/tex]
:3

Answer Link

joaobezerra joaobezerra · Answer 2 · 2019-12-26T17:50:56+01:00

Answer:

The maximum height attained by the projectile is 412.94 meters = 1359.79 feet.

Step-by-step explanation:

The height formula has the following formula:

[tex]h = h_{0} + v_{0}t - \frac{gt^{2}}{2}[/tex]

In which [tex]h_{0}[/tex] is the initial height, [tex]v_{0}[/tex] is the initial speed and g = 9.8 m/s² is the gravity.

In this problem, we have that:

We are working with the gravity in meters, so we must convert the feet measures to meters.

Each feet has 0.3048 meters.

So 288 feet per second = 87.78 meters per second.

65 foot = 19.81 meters.

This means that:

[tex]h_{0} = 19.81, v_{0} = 87.78[/tex]

So

[tex]h(t) = 19.81 + 87.78t - 4.9t^{2}[/tex]

The maximum height is attained at the moment of time in which the velocity is 0. The velocity is the derivative of the height. So:

[tex]v(t) = h'(t) = -9.8t + 87.78[/tex]

[tex]v(t) = 0[/tex]

[tex]9.8t = 87.78[/tex]

[tex]t = 8.96[/tex]

The maximum height is attained at 8.96s. This height is

[tex]h(t) = 19.81 + 87.78t - 4.9t^{2}[/tex]

[tex]h(8.96) = 19.81 + 87.78*8.96 - 4.9*(8.96)^{2} = 412.94[/tex]

The maximum height attained by the projectile is 412.94 meters = 1359.79 feet.