A ball is thrown vertically upward with an initial speed of 25 m/s. Then, 1.8 s later, a stone is thrown straight up (from the same initial height as the ball) with an initial speed of 32.1 m/s .
How far above the release point will the ball and stone pass each other? The acceleration of gravity is 9.8 m/s2 .
Answer in units of m.
x1 = 25(t) - 4.9t^2
x2 = 32.1 (t+1.8) - 4.9(t+1.8)^2
Set them equal to each other. Solve for t (note t=zero is a trivial solution), plug the t back into the originals.
To determine the distance above the release point where the ball and stone pass each other, we can calculate the heights reached by each object at the specific time.
First, let's calculate the height reached by the ball after 1.8 seconds.
The formula to determine the height of an object thrown vertically upward is:
h = v₀t - (1/2)gt²
Where:
h = height
v₀ = initial velocity
t = time
g = acceleration due to gravity
Given that the initial velocity of the ball is 25 m/s, and the time is 1.8 seconds, we can substitute these values into the formula:
h(ball) = (25 m/s)(1.8 s) - (1/2)(9.8 m/s²)(1.8 s)²
Now, let's calculate the height reached by the stone at the same time.
Using the same formula, we can substitute the initial velocity of the stone (32.1 m/s) and the time (1.8 s):
h(stone) = (32.1 m/s)(1.8 s) - (1/2)(9.8 m/s²)(1.8 s)²
To determine the distance between the ball and stone at the point of passing, we need to subtract the height reached by the stone from the height reached by the ball:
Distance = h(stone) - h(ball)
Now we can substitute the calculated values into the formula and solve for the distance:
Distance = [(32.1 m/s)(1.8 s) - (1/2)(9.8 m/s²)(1.8 s)²] - [(25 m/s)(1.8 s) - (1/2)(9.8 m/s²)(1.8 s)²]
To find the distance above the release point where the ball and stone pass each other, we need to determine the heights reached by both objects at the time of their meeting.
First, let's calculate the maximum height reached by the ball. The ball is thrown vertically upward, so its motion can be modeled by the equation:
h_ball = h_initial + v_initial*t - 0.5*g*t^2
Where:
h_ball is the height of the ball,
h_initial is the initial height of the ball,
v_initial is the initial velocity of the ball,
g is the acceleration due to gravity (9.8 m/s^2),
t is the time since the ball was thrown.
Given:
h_initial = 0 (since we are measuring heights from the release point)
v_initial_ball = 25 m/s
t_ball = t (since the ball is released first)
Plugging in the values, we get:
h_ball = 0 + 25*t_ball - 0.5*(9.8)*t_ball^2
Next, let's calculate the maximum height reached by the stone. The stone is thrown straight up, so its motion can also be modeled by the same equation:
h_stone = h_initial + v_initial_stone*t - 0.5*g*t^2
Where:
h_stone is the height of the stone,
v_initial_stone is the initial velocity of the stone,
t_stone is the time since the stone was thrown.
Given:
h_initial_stone = 0 (since we are measuring heights from the release point)
v_initial_stone = 32.1 m/s
t_stone = t_ball + 1.8 s (since the stone is released 1.8 seconds after the ball)
Plugging in the values, we get:
h_stone = 0 + 32.1*(t_ball + 1.8) - 0.5*(9.8)*(t_ball + 1.8)^2
Now, to find the distance above the release point where the ball and stone pass each other, we need to find the time at which their heights are equal. We can set h_ball equal to h_stone and solve for t_ball:
25*t_ball - 0.5*(9.8)*t_ball^2 = 32.1*(t_ball + 1.8) - 0.5*(9.8)*(t_ball + 1.8)^2
Simplifying and solving this equation will give us the value of t_ball. Substituting this value into the equation for h_ball will give us the height at which the ball and stone pass each other, which is the distance above the release point.