A light is at the top of a pole 80 feet high. A ball is dropped at the same
height from a point 20 feet away from the light. A wall 80 feet high, 60 feet
away from the light is built. Assuming the ball falls according to the
Newtonian Law 2 s=16(t^2) where s is the distance in feet and t is the time in
seconds, find:
a) how fast the shadow of the ball is moving on the wall after 1 second.
b) how fast the shadow is moving along the ground after 2 seconds.
a. -100
To find the speed at which the shadow of the ball moves on the wall and along the ground, we need to first determine the positions of the ball and its shadow at different points in time.
Let's start by finding the time it takes for the ball to hit the ground. We can do this by setting the position of the ball, s, equal to the height of the pole, 80 feet, and solving for t:
80 = 16(t^2)
(t^2) = 5
Taking the square root of both sides, we find:
t = √5
a) To find how fast the shadow of the ball is moving on the wall after 1 second, we need to find the position of the ball on the wall after 1 second.
Substituting t = 1 into the equation s = 16(t^2), we get:
s = 16(1^2) = 16
Therefore, after 1 second, the ball has fallen 16 feet. Since the ball is dropped 20 feet away from the light, the shadow of the ball on the wall is at a position 4 feet away from the light (20 - 16 = 4).
The speed of the shadow on the wall is the rate of change of its position with respect to time, which is given by ds/dt.
ds/dt = d(4)/dt = 0
The shadow of the ball on the wall is not moving at all after 1 second.
b) To find how fast the shadow is moving along the ground after 2 seconds, we need to find the position of the ball on the ground after 2 seconds.
Substituting t = 2 into the equation s = 16(t^2), we get:
s = 16(2^2) = 64
Therefore, after 2 seconds, the ball has fallen 64 feet. The distance between the ball's starting point (20 feet away from the light) and the point where it hits the ground is 80 - 64 = 16 feet.
The speed of the shadow along the ground is the rate of change of its position with respect to time, which is given by ds/dt.
ds/dt = d(16)/dt = 0
The shadow of the ball along the ground is not moving at all after 2 seconds.
Therefore, after 1 second, the shadow of the ball on the wall is not moving, and after 2 seconds, the shadow of the ball along the ground is also not moving.
To find the speed of the shadow of the ball on the wall and the ground, we need to find the rates of change of the horizontal and vertical distances.
a) To find the speed of the shadow on the wall after 1 second:
According to the given information, the ball is dropped from a point 20 feet away from the light. Therefore, we can consider a right triangle formed by the light pole, the wall, and the shadow of the ball.
Let's define the vertical distance as y, the horizontal distance as x, and the time elapsed as t.
Using the Pythagorean theorem:
x^2 + y^2 = 80^2
Differentiating both sides with respect to time t:
2x(dx/dt) + 2y(dy/dt) = 0
(dx/dt) = -y/(x(dy/dt))
Using the given equation 2s = 16t^2:
s = 8t^2
Differentiating both sides with respect to time t:
(ds/dt) = 16t
Since s is the horizontal distance x:
(dx/dt) = 16t
To find the values of x and y at t = 1:
x = 20 - 16(1)
x = 20 - 16 = 4 ft
y = sqrt(80^2 - 4^2)
y = sqrt(6296) ≈ 79.36 ft
Substituting the values of x and y into the equation (dx/dt) = 16t:
(dx/dt) = 16(1)
(dx/dt) = 16 ft/s
Therefore, the shadow of the ball on the wall is moving at a speed of 16 ft/s after 1 second.
b) To find the speed of the shadow on the ground after 2 seconds:
Using the same approach as above, we can find the horizontal distance x and the vertical distance y at t = 2:
x = 20 - 16(2)
x = 20 - 32 = -12 ft
y = sqrt(80^2 - (-12)^2)
y = sqrt(6344) ≈ 79.64 ft
Substituting the values of x and y into the equation (dx/dt) = 16t:
(dx/dt) = 16(2)
(dx/dt) = 32 ft/s
Therefore, the shadow of the ball on the ground is moving at a speed of 32 ft/s after 2 seconds.