A camera is set up to take photographs of a ball
undergoing vertical motion. The camera is 5.2 m
above the ball launcher, a device that can launch the
ball with an initial velocity 17 m/s [up]. Assuming
that the ball goes straight up and then straight down
past the camera, at what times after the launch will
the ball pass the camera?
To find the time it takes for the ball to pass the camera, we need to consider the motion of the ball as it goes up and then comes back down.
When the ball is at its highest point, its velocity will be 0 m/s. This is because at the highest point of its trajectory, the ball will momentarily stop before falling back down due to gravity.
We can use the equation of motion for vertical motion:
v = u + at
Where:
v = final velocity (m/s)
u = initial velocity (m/s)
a = acceleration due to gravity, approximately -9.81 m/s^2
t = time (s)
When the ball reaches its highest point, v = 0. So:
0 = 17 - 9.81t
9.81t = 17
t = 17 / 9.81
t ≈ 1.73 s
Therefore, it takes approximately 1.73 seconds for the ball to reach its highest point.
To find the total time it takes for the ball to pass the camera, we need to consider the entire time of flight. The time taken to reach the highest point is the same as the time taken for the ball to fall back down and reach the camera.
Thus, the ball will pass the camera approximately 1.73 seconds after launch and again when it comes back down to the same height as the camera.