A ball is thrown vertically upward with an initial speed of 11 m/s . Then, 0.67 s later, a stone is thrown straight up (from the same initial height as the ball) with an initial speed of 25 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
To find out how far above the release point the ball and stone will pass each other, we need to determine the heights each object reaches and then subtract their initial height.
To determine the height of the ball, we can use the following kinematic equation:
h_ball = h_0 + v_0*t - (1/2)*g*t^2
where:
h_ball is the height of the ball,
h_0 is the initial height (assumed to be 0 in this case),
v_0 is the initial velocity of the ball (11 m/s),
g is the acceleration due to gravity (9.8 m/s^2),
t is the time.
Given that the ball is thrown 0.67 seconds before the stone, we can use that time (t_ball) to calculate the height of the ball. To find the height of the stone, we will use the same equation, but with a different initial velocity.
To calculate the height of the stone, we use:
h_stone = h_0 + v_0*t_stone - (1/2)*g*t_stone^2
where:
h_stone is the height of the stone,
t_stone is the time the stone is in the air (which is t_ball + 0.67 s),
v_0 is the initial velocity of the stone (25 m/s).
Now, let's calculate the heights of the ball and the stone:
For the ball:
h_ball = 0 + 11*0.67 - (1/2)*9.8*(0.67)^2
h_ball = 3.7 meters (rounded to one decimal place)
For the stone:
h_stone = 0 + 25*(0.67 + 0.67) - (1/2)*9.8*(0.67 + 0.67)^2
h_stone = 17.6 meters (rounded to one decimal place)
Finally, to find the distance above the release point where the ball and stone pass each other, we subtract the initial height (0 meters) from the difference in their heights:
Distance = h_stone - h_ball
Distance = 17.6 - 3.7
Distance = 13.9 meters
Therefore, the ball and stone will pass each other 13.9 meters above the release point.