A street light is mounted at the top of a 15-ft-tall pole. A man 6 ft tall walks away from the pole with a speed of 7 ft/s along
a straight path. How fast is the tip of his shadow moving when he is 30 ft from the pole?
2 answers:
Answer:
Since only his speed affects how fast his shadow moves, then we calculate for the speed which is 11.67ft/s
Step-by-step explanation:
Assuming similar triangle
(y-x)/y = 6/15
15(y-x) = 6y
15y - 15x = 6y
(15 -6)y = 15x
9y = 15x
y = (5/3)x
By differentiating both sides with respect to time t
dy/dt = 5/3 (dx/dt)
But (dx/dt) is 7 ft/s
Therefore dy/dt = 5/3 * 7 = 35/3 = 11.67ft/s
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Answer:
12.7m
Step-by-step explanation:
The equation for motion to be used to find how far the water travels is
xf = xi + Vix(t) + 1/2axt^2
xf = 0 + ViCos α(t) + 0
xf = ViCosα(t)
xf = 12 Cos 30°(t)
xf = (6√3)t
To find the total times we will use the vertical displacement equation
yf = yi + Viy(t) + 1/2ayt^2
0 = 0 + ViSinα(t) + (1/2 * (-9.8)t^2)
0 = 12Sin30°(t) - 4.9t^2
4.9t^2 = 12Sin30°(t)
4.9t^2 = 6t
Divide through by t
4.9t = 6
t = 6/4.9
t = 1.22secs
Put t = 1.22 into (6√3)t
xf = 6√3 * 1.22
xf = 12.7m
