Question

Michael is standing on top of a building when he drops a lit smoke bomb. How far down does the smoke bomb detonate if it falls for 3.7 seconds?

Answer 1

The smoke bomb would fall approximately 67.22 meters in 3.7 seconds due to free fall before detonating.

Step-by-step explanation:

To determine how far down the smoke bomb detonates if it falls for 3.7 seconds, we need to use the formula for the distance traveled during free fall, which is d = ½ g t^2. The variable d is the distance, g is the acceleration due to gravity (approximately 9.81 m/s2 on Earth), and t is the time the object has been falling.

By inserting the given time of 3.7 seconds, we calculate:

d = ½ * 9.81 m/s2 * (3.7 s)2
= 67.22 m

Therefore, the smoke bomb would detonate approximately 67.22 meters below the point of release.