95.5k views
3 votes
You stand on a bridge and drop a teddy bear. The teddy bear hits the water below the bridge in 3.5 seconds. One second after dropping the teddy bear, you decide to toss your rubber ducky down as well. How fast would you have to throw the rubber ducky so it hits the water at the same time? physics

User Sybreon
by
7.7k points

1 Answer

4 votes

To make the rubber ducky hit the water at the same time as the teddy bear that was dropped from a bridge, you would need to throw it downwards with an initial velocity of 24.01 m/s, one second after the bear was dropped.

To determine how fast you would need to throw the rubber ducky to ensure it hits the water at the same time as the teddy bear, we'll use the equations of motion under constant acceleration due to gravity, assuming negligible air resistance.

First, we calculate the height from which the objects are dropped using the time it took for the teddy bear to hit the water.

The equation for the distance fallen under gravity is:

s = ½ gt²

Where s is the distance fallen, g is the acceleration due to gravity (approximately 9.80 m/s²), and t is the time in seconds.

Plugging in the teddy bear's fall time:

s = ½ * 9.80 m/s² * (3.5 s)² = 60.025 m

The rubber ducky is thrown 1 second after the teddy, so it has 2.5 seconds to hit the water. Ignoring air resistance, the initial velocity v needed for the ducky can be found using the equation:

v = s / t - ½ g t

Substituting in the values:

v = 60.025 m / 2.5 s - ½ * 9.80 m/s² * 2.5 s = 24.01 m/s

Thus, to hit the water simultaneously with the teddy bear, you would need to throw the rubber ducky downwards at 24.01 m/s.

User Donel
by
8.0k points