A ball is thrown straight up from the edge of the roof of a 60.0 m tall building. If a second ball is
dropped from the roof 2.0 s later what must the initial velocity of the first ball be if both are to reach
the ground at the same time?
The second ball hits when
60-4.9(t-2)^2 = 0,
t = 5.5
v = 19.6(1 - 1/5.5) = 10.036 m/s
So, you want v where
60 + 5.5v - 4.9*5.5^2 = 0
v = 16.04 m/s
check:
To solve this problem, we can use the equations of motion for both balls. Let's denote the initial velocity of the first ball as "v1" and the acceleration due to gravity as "g" (approximately 9.8 m/s^2).
For the first ball:
- The initial velocity is 'v1'.
- The acceleration is -g (since the ball is moving against gravity).
- The distance it needs to travel is the height of the building, which is 60.0 m.
- The time it takes for the ball to reach the ground can be found by setting the displacement equal to zero in the equation of motion for vertical motion: s = ut + (1/2)at^2. Since the ball reaches the ground, the displacement is 0. Hence, we have 0 = v1t + (1/2)(-g)t^2.
For the second ball:
- The initial velocity is 0 m/s (since it is dropped).
- The acceleration is g (since it is moving with gravity).
- The distance it needs to travel is also the height of the building, which is 60.0 m.
- The time it takes for the ball to reach the ground can be found using the equation of motion: s = ut + (1/2)at^2. In this case, we know the initial velocity is 0, so the equation simplifies to 60 = (1/2)(g)t^2.
Now, we can solve for 't' in both equations and equate them to find the initial velocity of the first ball.
- For the first ball, the equation is 0 = v1t - (1/2)gt^2.
- For the second ball, the equation is 60 = (1/2)(g)t^2.
Since both balls reach the ground at the same time, the time taken by both balls must be the same.
By equating the two equations, we get:
v1t - (1/2)gt^2 = (1/2)(g)t^2
v1t = (1/2)gt^2 + (1/2)gt^2
v1t = gt^2
v1 = gt
Substituting the value of 't' from the second ball equation:
v1 = g * (2) = 9.8 * 2 = 19.6 m/s
Therefore, the initial velocity of the first ball (v1) must be 19.6 m/s to ensure that both balls reach the ground at the same time.