The height above the ground of an object thrown upward from a point feet above the ground with an initial velocity of feet per second is given by the function $f ( t ) = - 16 t ^ { 2 } + v _ { 0 } t + s _ { 0 }$. A balloon, rising vertically with a velocity of 24 feet per second, releases a sandbag at the instant it is 20 feet above the ground. At what velocity will it hit the ground? Round your answer to three decimal places.
A) $- 60.111 \mathrm { ft } / \mathrm { sec }$
B) $- 43.081 \mathrm { ft } / \mathrm { sec }$
C) $- 15.028 \mathrm { ft } / \mathrm { sec }$
D) $49.298 \mathrm { ft } / \mathrm { sec }$
E) $67.081 \mathrm { ft } / \mathrm { sec }$

Question

Quizplus · Accepted Answer

The initial velocity $v_0$ of the sandbag is 24 ft/s (same as the balloon). Using the formula for velocity $v(t) = -32t + v_0$, we find the velocity when it hits the ground. Solving for $t$ when $f(t) = 0$ gives the time of impact, and substituting back into $v(t)$ gives the velocity.

The Height Above the Ground of an Object Thrown Upward f(t)=−16t2+v0t+s0f ( t ) = - 16 t ^ { 2 } + v _ { 0 } t + s _ { 0 }f(t)=−16t2+v0​t+s0​

The Height Above the Ground of an Object Thrown Upward $f ( t ) = - 16 t ^ { 2 } + v _ { 0 } t + s _ { 0 }$