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