A street light (light source) is 15 m above the ground. A ball is freely released from a location of 9 m from the light with the same elevation as the light’s. While the ball is falling, it casts a shadow on the ground. Assuming the gravity is 10 m/s2, calculate the speed of the shadow of the ball at 0.5 s after the ball has been released.
the height y of the ball at time t is
y = 15 - 5t^2
The shadow's position x meters from where the ball will hit, is found by using similar triangles:
y/x = 15/(9+x)
(15 - 5t^2)/x = 15/(9+x)
dx/dt = -(9+x)/t
at t = .5, x = 99, so
dx/dt = -(9+99)/(.5) = -216 m/s
To calculate the speed of the shadow of the ball at 0.5 seconds after it has been released, we can consider the motion of the ball and the shadow separately.
First, let's calculate the time it takes for the ball to fall from its initial position to a height of 0 meters (i.e., when it reaches the ground). We can use the equation of motion:
h = ut + (1/2)gt^2
where h is the displacement (change in height), u is the initial velocity (0 since the ball is released from rest), g is the acceleration due to gravity (10 m/s^2), and t is the time.
Substituting the known values:
0 = 0 + 0.5 * 10 * t^2
Simplifying the equation:
5t^2 = t^2 = 0
t^2 = 0.5
t = √0.5 ≈ 0.71 seconds
So, it takes approximately 0.71 seconds for the ball to fall to the ground.
Now, let's calculate the distance the shadow travels during this time. Since the shadow is cast along a straight line on the ground, it travels the same horizontal distance as the ball. From the given information, we know that the ball is released from a distance of 9 meters from the light.
Therefore, the distance covered by the shadow during the 0.71 seconds is 9 meters.
To find the speed of the shadow, we divide the distance by time:
Speed = Distance / Time
Speed = 9 meters / 0.71 seconds
Speed ≈ 12.68 meters per second
Hence, the speed of the shadow of the ball at 0.5 seconds after the ball has been released is approximately 12.68 meters per second.