Question

A boy projects a ball up with an initial velocity of 80 ft/s. The ball will be at a height of 96 ft from the ground after

• A. 2 s
• B. 3 s
• C. 5 s
• D. both (a) and (b)

Answer 1

The correct option is D both (a) and (b)

From equation of motion, we have
$s=ut-\frac{1}{2}g{t}^2$
Given, s=96 ft, u= 80 ft/s, g = 32 $ft/s^2$
$\therefore 96=80t-\frac{32}{2}{t}^2$
$\Rightarrow t^2-5t+6=0$
$\Rightarrow t^2-3t-2t+6=0$
$\Rightarrow t(t-3)-2(t-3)=0$
$\Rightarrow (t-2)(t-3)=0$
$\Rightarrow t=2or3s$