A ball is dropped from rest from a height of 20.0m. One second later a second ball is thrown vertically downwards. If the two balls arrive on the ground at the same time, what must have been the initial velocity of the second ball?

Question

A ball is dropped from rest from a height of 20.0m. One second later a second ball is thrown vertically downwards. If the two balls arrive on the ground at the same time, what must have been the initial velocity of the second ball?

Answer 1

First Ball.

d = Vo*t + 0.5g*t^2 = 20 m.
0 + 4.9t^2 = 20
t^2 = 4.08
Tf = 2.02 s. = Fall time

2nd Ball.
d = Vo*t + 0.5g*T^2 = 20.
Vo*(2.02-1) * 4.9*(2.02-1)^2 = 20
1.02Vo + 4.9*(1.02)^2 = 20
1.02Vo + 5.1 = 20
1.02Vo = 20 - 5.1 = 14.9
Vo = 14.6 m/s.

Answer 2

To solve this problem, we can use the kinematic equations of motion. Let's assume that the downward direction is positive.

For the first ball:

Initial position (s_1) = 20.0 m
Final position (s_2) = 0 m
Initial velocity (u_1) = 0 m/s (dropped from rest)
Final velocity (v_1) = ?
Time (t_1) = t seconds

Using the equation:
v_1 = u_1 + gt_1

Since the ball is dropped from rest, the initial velocity is 0, so we get:
v_1 = 0 + 9.8t_1
v_1 = 9.8t_1

For the second ball:

Initial position (s_3) = 0 m
Final position (s_4) = 0 m
Initial velocity (u_2) = ? (We need to find this)
Final velocity (v_2) = ?
Time (t_2) = t - 1 seconds (As the second ball is thrown 1 second later)

Using the equation:
v_2 = u_2 + gt_2

Since the final velocity is 0, we get:
0 = u_2 + 9.8(t - 1)
u_2 = -9.8t + 9.8

Now, we know that both balls reach the ground at the same time. Therefore, the time taken for both balls to reach the ground is equal. We can set up an equation using this information.

For the first ball: s_2 = 0 = 20 - (1/2)gt_1^2
For the second ball: s_4 = 0 = -0.5gt^2 + (u_2)(t-1)

From the first equation, we can find t_1:
20 = (1/2)gt_1^2
40/g = t_1^2
t_1 = sqrt(40/g)

Now, substitute t_1 into the second equation:
0 = -0.5gt^2 + (-9.8t + 9.8)(t - 1)

Simplifying and rearranging the equation, we get:
0 = -0.5gt^2 + 9.8t^2 - 19.6t + 9.8

Combining like terms, we get:
9.8t^2 - 19.6t + 9.8 - 0.5gt^2 = 0

Since both balls hit the ground at the same time, this equation can be solved to find the value of t. Substitute this value of t back into the equation for u_2 to find the initial velocity of the second ball.