A light pole of height 15 meter has a lamp hanging at its top . from a distance 12 meter from the base of the pole , a ball is thrown vertically upwards with velocity 5m/sec . find the rate at which the shadow of the ball on the earth moves away from the base of the pole after 0.5 second .
velocity ball=5-9.8t^2 upward....
draw the figure, if x is the distance from the pole to the shadown, one has two similar triangles
15/x=h/(x-12) where h is the height of the ball.
a) 15x-180=xh
but h=5t-1/2 9.8 t^2
dh/dt=5-9.8t
starting with the equation a)
15 dx/dt=hdx/dt + x dh/dt
or dx/dt (15-h)=x dh/dt
well, we know h as a function of t
and we know x :(x=180/(15-h):
dx/dt= x/(15-h) dh/dt
= 180/(15-h)^2 *(5-9.8t)
then put in for h (5t-1/2 9.8 t^2)
and you have it.
Thank you :)
To find the rate at which the shadow of the ball on the earth moves away from the base of the pole, we need to consider the similar triangles formed by the light pole, the shadow, and the ball.
Let's denote the distance of the ball from the base of the pole at time t as "d" (in meters). We are given that the ball is thrown vertically upwards with a velocity of 5m/sec. Therefore, at time t = 0.5 seconds, the distance of the ball from the base of the pole would be:
d = 12 + 5 * 0.5
d = 12 + 2.5
d = 14.5 meters
Now, let's consider the height of the light pole and the length of its shadow on the ground. The height of the light pole is given as 15 meters, and the distance from the base of the pole to the point where the shadow meets the ground is also 12 meters.
Using the similar triangles formed by the pole, its shadow, and the ball, we can establish the following proportion:
(height of pole) / (length of pole's shadow) = (distance of ball from pole) / (length of ball's shadow)
15 / x = 14.5 / 12
Cross-multiplying, we get:
15 * 12 = x * 14.5
180 = 14.5x
x = 180 / 14.5
x ≈ 12.4138 meters
Therefore, at time t = 0.5 seconds, the length of the ball's shadow on the ground is approximately 12.4138 meters.
To find the rate at which the shadow of the ball on the earth moves away from the base of the pole, we need to find the derivative of the length of the shadow with respect to time. Let's differentiate the equation 15 / x = 14.5 / 12 with respect to time:
d(15 / x) / dt = d(14.5 / 12) / dt
Using the quotient rule of differentiation, we get:
(-15 / x^2) * dx/dt = 0
Since the rate at which the shadow of the ball on the earth moves away from the base of the pole is given by dx/dt, we need to solve for that. Rearranging the equation, we get:
dx/dt = 0 / (-15 / x^2)
dx/dt = 0
Therefore, the rate at which the shadow of the ball on the earth moves away from the base of the pole after 0.5 seconds is 0 meters per second.