Question

A ball is shot from a cannon into the air with an upward velocity of 40 ft/sec. The equation that give the height (h) of the ball at any time (t) is: h(t)= -16t^2 + 40t + 1.5. Find the maximum height attained by the ball. I need a clear explanation because I have to expose this

Answer 1

The height of the ball at any time t is given by

`h(t)=-16t^2+40t+1.5`

This is a quadratic equation, which attains its maximum value at time:

`t=(-b)/(2a)`

In the given equation, a = -16 and b = 40. substitute these values in the formula:

`t=(-40)/(-16*2)=(-40)/(-32)=(5)/(4)`

Therefore, the ball attains its maximum height at t=5/4 seconds which is given below:

`\begin{gathered} h((5)/(4))=-16((5)/(4))^2+40((5)/(4))+1.5 \\ =-25+50+1.5 \\ =26.5 \end{gathered}`

Thus, the maximum height attained by the ball is 26.5 feet.