Question

a ball is thrown vertically upward with a velocity u. find the velocity when it touches the ground. class 9​

Answer 1

First, let's assume you are not taking air resistance into consideration. This is a reasonable assumption because it's obvious that a parachute's velocity just before it hits depends on factors other than just gravity or drop height. Adjusting its size can influence its speed at impact.

Without air friction, an object WILL regain its original upward velocity just before it hits the ground on the way down. The initial velocity on the ground represents KINETIC ENERGY, and the initial height of (0) represents NO POTENTIAL ENERGY. As the object rises, its kinetic energy is gradually converted into potential energy until at its highest point all the original kinetic energy (now zero) has been converted into potential energy. At any point in its travel, TOTAL kinetic and potential energy remains constant.

The process is reversed when the object falls. Starting at maximum potential energy (determined by its maximum height) and zero kinetic energy (not moving), the potential energy gradually gets converted back into kinetic energy until just before it hits the ground. Then, potential energy is zero and it has all the kinetic energy it had at the start.

So the CONSERVATION OF ENERGY is why the two velocities are the same.

then the other case is that it can be more or less than the original velocity.it depends on time.i think you know about the latter case.

hope it helps.

plz mark as brainliest.

Answer 2

Answer

Given,

Initial velocity, u=−μms−1​

Final velocity, v=3μms−1

Apply first kinematic equation of motion

 v=u+at

3μ=−μ+gt

t=g4μ​

Hence, time required to reach ground is g4μ​