A ball is thrown straight up from ground level. It passes a 2-m-high window. The bottom of the window is 7.5 m off the ground. The time that elapses from when the ball passes the bottom of the window on the way up, to when it passes the top of the window on the way back down is 1.3 s.
What was the ball’s initial speed, in meters per second?
First consider only the distance along the window, and solve for the ball’s velocity at the bottom of the window. Next, consider only the distance from the ground to the bottom of the window, and solve for the initial velocity using the velocity at the bottom of the window as the final velocity.
To find the ball's initial speed, we can use the kinematic equation that relates time, initial velocity, and displacement.
First, let's calculate the time it takes for the ball to pass the bottom of the window. We know that the bottom of the window is 7.5 m off the ground, and we can assume the window is at ground level. We can use the equation for displacement, s = ut + (1/2)at^2, where s is the displacement, u is the initial velocity, t is the time, and a is the acceleration (which we assume to be -9.8 m/s^2 due to gravity).
Since the ball is thrown straight up and then comes back down, the total displacement is the sum of the upward displacement and the downward displacement, which is zero. Let's first calculate the time it takes for the ball to reach the top of the window on its way up.
Using the equation s = ut + (1/2)at^2, we can substitute the known values:
0 = u(1.3) + (1/2)(-9.8)(1.3)^2
Simplifying the equation:
0 = 1.3u - 8.045
Rearranging the equation to solve for u:
1.3u = 8.045
u = 8.045 / 1.3
u ≈ 6.19 m/s
Therefore, the ball’s initial speed is approximately 6.19 meters per second.