Question

a ball is shot from a cannon into the air with an upward velocity of 40ft/sec. The equation that gives the height (h) of the ball at any time (t) is: h(t)= -15+30t+4

Answer 1

(a) Height at which the ball is shot from the canon

Write the equation of the height as a function of time.

`h(t)=-15t^2\text{ + 30t + 4}`

The initial height of the ball is the height at time t =0.

Hence:

`\begin{gathered} h(t=0)=-15(0)^2\text{ + 30(0) + 4} \\ =\text{ 4} \end{gathered}`

(b) Time at which the ball reaches the maximum height

We can obtain this by differentiating h(t) with respect to t, and then equating the expression to zero.

Taking the first derivative:

`h^(\prime)(t)\text{ = -30t + 30}`

Equating the zero:

`\begin{gathered} -30t\text{ +30 = 0} \\ -30t\text{ = -30} \\ \text{Divide both sides by -30} \\ (-30t)/(-30)=(-30)/(-30) \\ t\text{ = 1} \end{gathered}`

The ball reaches maximum height at t = 1s

(c) The maximum height the ball can reach

Substituting 1 for t into the expression h(t):

`\begin{gathered} h(t=1)=-15(1)^2\text{ +30(1) + 4} \\ =\text{ -15 + 30 + 4} \\ =\text{ 19} \end{gathered}`

The maximum height the ball can reach is 19ft

(d) How long for the ball to reach the ground

The ball would reach the ground at exactly 2 times the value it takes to reach the maximum height

Hence:

`\begin{gathered} \text{time = 2 }*1 \\ =\text{ 2s} \end{gathered}`

It would take 2s for the ball to reach the ground