A water balloon was dropped from a high window and struck its target 1.1 seconds later. If the balloon left the person's hand at -5.0 m/s, what was its velocity on impact?

To determine the water balloon's velocity on impact, you would substitute the initial velocity, time, and acceleration due to gravity into the kinematic equation for velocity, resulting in a final velocity of 5.78 m/s. ;

Answered by JeanneReese | 2024-06-18

The acceleration of gravity on Earth is 9.8 meters per second every second. So in 1.1 second of falling, an object gains (9.8 x 1.1) = 10.78 meters per second of downward speed.
The balloon was already falling at 5 m/sec when the time began, and it gained another 10.78 m/sec on the way. So at the end of 1.1 second, its downward speed was (5.0 + 10.78) = 15.78 m/sec.
Its velocity was 15.78 m/sec downward.

Answered by AL2006 | 2024-06-24

The velocity of the water balloon on impact is approximately 5.78 m/s downward. This is calculated using the initial velocity, the time of fall, and the acceleration due to gravity. The calculation shows how the initial downward velocity and the increase in speed due to gravity combine for the final impact velocity.
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Answered by AL2006 | 2024-09-30