A particle is projected up with an initial velocity of 80 ft/s. The ball will attain height of 96 ft/s from the ground after
A)2.0 and 3.0 s
B)Only at 3.0s
C)Only at 2.0 s
D)After 1 and 2 s

Question

Quizplus · Accepted Answer

The height $h$ of a projectile at time $t$ can be found using the formula $h = v_0t - \frac{1}{2}gt^2$, where $v_0$ is the initial velocity and $g$ is the acceleration due to gravity (32 ft/s$^2$ for Earth). Plugging in the given values, we get $96 = 80t - 16t^2$. Rearranging, we get $16t^2 - 80t + 96 = 0$. Factoring or using the quadratic formula, we find that the equation has two solutions for $t$, indicating that the ball reaches 96 ft at two different times, once on the way up and once on the way down, which corresponds to the times given in option A.

A Particle Is Projected Up with an Initial Velocity of 80