At time t = 0, a bowling ball rolls off a 250-meter ledge with velocity 30 meters/sec downward.Express its height, h(t), in meters above the ground as a function of time, t, in seconds.
A) $h(t)=-4.9 t^{2}-30 t+250$
B) $h(t)=-4.9 t^{2}-30 t-250$
C) $h(t)=4.9 t^{2}-30 t+250$
D) $h(t)=4.9 t^{2}-30 t-250$

Question

Quizplus · Accepted Answer

The correct formula for the height of an object in free fall is $h(t) = -\frac{1}{2}gt^2 + v_0t + h_0$, where $g$ is the acceleration due to gravity (approximately 9.8 m/s$^2$), $v_0$ is the initial velocity, and $h_0$ is the initial height. Substituting $g = 9.8$, $v_0 = -30$ (since the velocity is downward), and $h_0 = 250$, we get $h(t) = -4.9t^2 - 30t + 250$.

At Time T = 0, a Bowling Ball Rolls Off h(t)=−4.9t2−30t+250h(t)=-4.9 t^{2}-30 t+250h(t)=−4.9t2−30t+250

At Time T = 0, a Bowling Ball Rolls Off $h(t)=-4.9 t^{2}-30 t+250$