ll is thrown vertically upward with an
initial speed of 16 m/s. Then, 0.65 s later, a
stone is thrown straight up (from the same
initial height as the ball) with an initial speed
of 27 m/s.
How far above the release point will the ball
and stone pass each other? The acceleration
of gravity is 9.8 m/s
2
.
Answer in units of m
See previous post.
To find the distance above the release point where the ball and stone pass each other, we can determine the heights reached by each object at a given time.
First, let's calculate the time it takes for the ball and stone to reach their maximum heights.
For the ball:
Using the vertical motion equation:
v = u + at
Where:
v = final velocity (0 m/s, at maximum height)
u = initial velocity (16 m/s, upward)
a = acceleration (acceleration due to gravity, -9.8 m/s^2, downward)
Rearranging the equation to solve for time:
t = (v - u) / a
Substituting the values:
t_ball = (0 - 16) / -9.8
t_ball = -16 / -9.8
t_ball ≈ 1.63 s
For the stone:
Using the same process, we can calculate the time it takes for the stone to reach its maximum height. Since it is thrown 0.65 seconds after the ball, we subtract that time from the total:
t_stone = (0 - 27) / -9.8
t_stone = 27 / 9.8
t_stone ≈ 2.76 s
To find the heights reached by the ball and the stone at the time they pass each other, we use the formula:
s = ut + 0.5at^2
Where:
s = height
u = initial velocity
a = acceleration
t = time
For the ball:
s_ball = 16(1.63) + 0.5*(-9.8)(1.63)^2
For the stone:
s_stone = 27(2.11) + 0.5*(-9.8)(2.11)^2
Now, let's calculate:
s_ball = 26.08 m
s_stone = 66.79 m
Therefore, the ball and stone will pass each other at a height of approximately 26.08 meters above the release point.