Question

A ball with mass m projected horizontally off the end of a table with an initial kinetic energy k. At a time t after it leaves the end of the table it has kinetic energy 3k. What is t

Answer 1

Answer:

The time t after it leaves the end of the table it has kinetic energy 3K is

$t = \sqrt{\frac{4K}{mg^2}}$

Explanation:

Let the initial speed by which ball is projected is given as

$v = v_o$

now the kinetic energy of the ball is given as

$K_i = \frac{1}{2}mv_o^2$

now after time "t" when it is moving in air the speed of the ball will increase in Y direction

so final speed of the ball will be

$v_y = 0 + gt$

now we have final kinetic energy as

$K_f = \frac{1}{2}m(v_o^2 + (gt)^2)$

here we know that final kinetic energy is 3 times the initial energy

so we will have

$3(\frac{1}{2}mv_0^2) = \frac{1}{2}mv_0^2 + \frac{1}{2}m(gt)^2$

$3K = K + \frac{1}{2}mg^2t^2$

$t = \sqrt{\frac{4K}{mg^2}}$

#Learn

Topic : projectile motion

https://brainly.in/question/512838