Question

A projectile is launched from ground level with an initial velocity of v0 feet per second. Neglecting air resistance, its height in feet t seconds after launch is given by s=-16t^2 +v0t. Find the time(s) that the projectile with (a) reach a height of 64 ft and (b) return to the ground when v0 = 80 feet per second

Answer 1

The formula is
`s=-16t^2+V_ot`

We have:

`s=64`

`V_o=80ft/s`

Substituting these values into the formula, we have


`64=-16t^2+80t`


`16t^2-80t+64=0` ⇒ divide each term by 16


`t^2-5t+4=0` ⇒ find the pair of numbers that multiply to give 4 and sum gives -5 which is -1 and -4


`(t-1)(t-4)=0` ⇒ We have two values of t


`t-1=0` ⇒
`t=1`

`t-4=0` ⇒
`t=4`

The time when the projectile reached 64 feet is at t=1 and t=4