Question

A projectile is launched from ground level with an initial velocity of v0 feet per second. Neglecting air resistance, it’s height in feet t seconds after launch is given by s= -16t^2+v0t. Find the time(s) that the projectile will (a) reach a height of 272 ft and (b) return to the ground when v0 is 32 feet per second

Answer 1

We know that the inital velity for this motion is 32 ft/s, this means that the height is given by:

`s=-16t^2+32t`

a)

To determine at what time the projectile will reach 272 we plug s=272 in the equation above and solve for t:

`\begin{gathered} -16t^2+32t=272 \\ 16t^2-32t+272=0 \\ t^2-2t+17=0 \\ \text{ then} \\ t=\frac{-(-2)\pm\sqrt[]{(-2)^2-4(1)(17)}}{2(1)} \\ t=\frac{2\pm\sqrt[]{-64}}{2} \end{gathered}`

since the discrimant of the equation is negative this means that the equation does not have real solutions.

This means that the projectile does not reach that height.

b)

To find at what time the projectile return to the ground we equate the expression for the height to zero and solve for t:

`\begin{gathered} -16t^2+32t=0 \\ t(-16t+32)=0 \\ \text{then} \\ t=0 \\ or \\ -16t+32=0 \\ 16t=32 \\ t=(32)/(16) \\ t=2 \end{gathered}`

Therefore it takes 2 seconds for the projectile to return to the ground