Question

a cannonball is fired directly upward with an initial velocity of 540 feet per second. its height above the ground at time t can be modeled with the equation

( h = - 16 t ^ { 2 } + 540 t )

how high does the cannonball travel before it begins to fall back to the ground? round your answer to two decimal places if needed.

height
135
ft
preview
135 ft

Explanation:

Step1: Find the time when the velocity is 0

The velocity function \(v(t)\) is the derivative of the height function \(h(t)=-16t^{2}+540t\). Using the power rule \((x^{n})^\prime = nx^{n - 1}\), we have \(v(t)=h^\prime(t)=-32t + 540\). Set \(v(t)=0\) (at the maximum height, velocity is 0):
\(-32t+540 = 0\)
\(32t=540\)
\(t=\frac{540}{32}=\frac{135}{8}=16.875\)

Step2: Substitute the time into the height function

Substitute \(t = 16.875\) into \(h(t)=-16t^{2}+540t\):
\(h(16.875)=-16\times(16.875)^{2}+540\times16.875\)
First, calculate \((16.875)^{2}=284.765625\)
\(-16\times284.765625=-4556.25\)
\(540\times16.875 = 9112.5\)
\(h(16.875)=- 4556.25+9112.5\)
\(h(16.875)=4556.25\)

Answer:

\(4556.25\)