Question

Two balls A and B of masses Ma and Mb are pushed horizontally from a table of height h.Ball A is pushed so that its intially velocity is vA and ball B is pushed so that its intially velocity is vB.Find the time it takes each ball to hit the ground.The acceleration due to gravity is g.​

Answer 1

The time taken by both balls to hit the ground is same and equal to
`\sqrt{(2h)/(g)}`

Step-by-step explanation:

Both the balls are pushed with a horizontal initial velocity. Therefore, the vertical component of initial velocity for both the balls is zero as the balls only have horizontal component of initial velocity.

Now, consider the equation of motion in the vertical direction for each ball.

Consider the upward direction as positive and downward direction as negative.

Ball A:

Given:

Initial vertical velocity is,
`u_y=0`

The displacement of the ball A is,
`y-y_0=0-h=-h`

Acceleration in the vertical direction is due to gravity,
`a_y=-g`

Therefore, we have the equation of motion as:

`y-y_0=u_yt_A+(1)/(2)a_yt_A^2\\-h=0+(1)/(2)(-g)t_A^2\\t_A^2=(2h)/(g)\\t_A=\sqrt{(2h)/(g)}`

Ball B:

Given:

Initial vertical velocity is,
`u_y=0`

The displacement of the ball A is,
`y-y_0=0-h=-h`

Acceleration in the vertical direction is due to gravity,
`a_y=-g`

Therefore, we have the equation of motion as:

`y-y_0=u_yt_B+(1)/(2)a_yt_B^2\\-h=0+(1)/(2)(-g)t_B^2\\t_B^2=(2h)/(g)\\t_B=\sqrt{(2h)/(g)}`

Therefore, the time taken by both the balls is same and equal to
`\sqrt{(2h)/(g)}`