Question

Instead of handing out candy like a well-adjusted person on Halloween, grumpy, anti-social Professor Sosa decides to stand on top of a 400 ft building and toss candy treats onto the ground below. If he drops a bag of candy corn, how long will it take for it to land on the ground? What will the velocity of the bag be when it hits the ground?

Answer 1

Answer: The bag of candy corn will take 5 seconds to land on the floor and it will hit it with velocity of 49m/s.

Step-by-step explanation: A free falling object has only Gravitational Force acting on it, so acceleration will be due to gravity, which has magnitude

g = 9.8m/s².

As Professor Sosa drops the bag, its initial velocity and time are 0.

The building has a height of 400ft, so adjusting units: y = 121.92 m and $y_{0}$ = 0.

For this problem, going down will be the positive reference.

Then:

$y=y_{0}+v_{0}t+\frac{g}{2}t^{2}$

$\Delta y=v_{0}t+\frac{g}{2}t^{2}$

$121.92=\frac{9.8}{2}t^{2}$

$t=\sqrt{\frac{121.92*2}{9.8} }$

t ≈ 5

It will take 5s for the candy corn bag to land on the ground.

Acceleration is the change in velocity per change in time, so:

$g=\frac{\Delta v}{\Delta t}$

As initial velocity and time are 0 and rearraging it:

v = g.t

v = 9.8*5

v = 49

The bag will hit the ground with velocity of magnitude 49 m/s.