Question

a pool ball leaves a 0.60 meter high table with an initial high table with an initial horizontal velocity of 2.4m/s. predict the time required for the pool ball to fall to the ground and the horizontal distance between the tables edge and the balls landing location

Answer 1

Ball leaves the table of height 0.60 m

Initial speed of the ball is in horizontal direction which is 2.4 m/s

Now here we can say that time taken by the ball to hit the ground will be given by kinematics equation in vertical direction

$\Delta y = v_y*t + \frac{1}{2}at^2$

here in y direction we can say

$\Delta y = 0.60 m$

$v_y = 0$

$a = 9.8 m/s^2$ due to gravity

now from above equation we have

$0.60 = 0 + \frac{1}{2}*9.8*t^2$

$t = 0.35 s$

Now in the same time the distance that ball move in horizontal direction is given as

$d = v_x* t$

given that

$v_x = 2.4 m/s$

$d = 2.4 * 0.35$

$d = 0.84 m$

So it will move a total distance of 0.84 m where it will land