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
h1 = Vo*t + 0.5g*t^2
h1 = 16*0.65 - 4.9*0.65^2 = 8.33 m. Head
start.
V1 = Vo + g*t = 16 - 9.8*0.65=9.63 m/s.
h2 = Vo*t + 0.5g*t^2=8.33+V1*t+0.5g*t^2
27*t - 4.9t^2 = 8.33 + 9.63t - 4.9t^2
27t - 9.63t - 4.9t^2 + 4.9t^2 = 8.33
17.37t = 8.33
t = 0.48 s.
h = 27*0.48 - 4.9*0.48^2 = 11.83 m Above
launching point.
To find the distance above the release point where the ball and stone pass each other, we need to determine the heights reached by each object at the time they cross paths.
First, let's calculate the height reached by the ball after 0.65 seconds. We can use the following kinematic equation:
h_ball = h_initial + v_initial * t + (1/2) * g * t^2
where:
h_ball = height reached by the ball
h_initial = initial height of the ball (same as the stone)
v_initial = initial velocity of the ball (16 m/s)
t = time taken by the ball to reach this height (0.65 seconds)
g = acceleration due to gravity (9.8 m/s^2)
Substituting the known values into the equation, we get:
h_ball = 0 + 16 * 0.65 + (1/2) * 9.8 * (0.65)^2
Simplifying the equation:
h_ball = 10.4 + 3.185
h_ball = 13.6 meters (rounded to one decimal place)
Now, let's calculate the height reached by the stone. The stone is thrown straight up, so the equation remains the same:
h_stone = h_initial + v_initial * t + (1/2) * g * t^2
Substituting the known values into the equation, we get:
h_stone = 0 + 27 * 0.65 + (1/2) * 9.8 * (0.65)^2
Simplifying the equation:
h_stone = 17.55 + 3.185
h_stone = 20.7 meters (rounded to one decimal place)
Therefore, the ball and stone pass each other at a height of 13.6 meters above the release point.