A diver jumps off a diving board with a speed of 25m/s he returns to the water after 3s. At the top of his dive describe his velocity and his acceleration?
To describe the velocity and acceleration of the diver at the top of his dive, we first need to understand the motion of the diver.
Given that the diver jumps off the diving board with a speed of 25 m/s and returns to the water after 3 seconds, we can assume that the motion of the diver follows a parabolic path because of gravity.
At the top of the diver's dive, the velocity is zero. This is because the moment the diver reaches the highest point, the velocity reaches its maximum value and starts to decrease. The velocity is constantly changing due to the acceleration of gravity. Therefore, at the top of the dive, the velocity is momentarily zero.
Regarding the acceleration at the top of the dive, we know that gravity is acting continuously on the diver, causing a constant acceleration downward. The acceleration due to gravity near the Earth's surface is approximately 9.8 m/s². Thus, at the top of the dive, the diver's acceleration will be approximately -9.8 m/s² (negative because it is directed downwards).
To summarize:
- At the top of the dive, the velocity is momentarily zero.
- The acceleration at the top of the dive is approximately -9.8 m/s².