Question 1126064: Please help. FREE FALL. A girl throws a ball vertically upward with a speed of 20 ft/s from the roof of a building 60 ft high. How long will it take the ball to reach the ground? What will be its speed when it strikes the ground?
Answer by rothauserc(4718) (Show Source):
You can put this solution on YOUR website! a) To find the time it takes to reach the ground, we solve this equation
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s(t) = -16t^2 +v(0)t +h, when s(t)=0, v(0)=20, h=60
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-16t^2 +20t +60 = 0
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divide both sides of = by -4
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4t^2 -5t -15 = 0
use the quadratic formula to solve
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t = (-(-5) +square root((-5)^2 -4*(4)(-15)))/(2*4) = 2.6599
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t = (-(-5) -square root((-5)^2 -4*(4)(-15)))/(2*4) = -1.4099
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it takes the ball 2.6599 seconds to reach the ground
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b) Due to symmetry, the ball will be moving 20 ft/sec when it comes back down to the point from which you threw it on the building. So you can reduce the problem to describing a ball thrown off a building with initial velocity 20 ft/sec downward.
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The balls position respect to time is then,
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s(t) = 60 -20t -16t^2
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to find t when s = 0, solve
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-16t^2 -20t +60 = 0
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divide both sides of = by -4
:
4t^2 +5t -15 = 0
:
use quadratic formula to solve for t
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t = (-5 +square root(5^2 -4*(4)(-15)))/(2*4) = 1.4099
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t = (-5 -square root(5^2 -4*(4)(-15)))/(2*4) = -2.6599
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we reject the negative time value
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t = 1.4099 seconds
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since acceleration is constant for this problem, we have
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v(t) = -20 -32t
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v(0.7352) = -20 -32(1.4099) = −65.1168
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The negative sign comes from calling the upwards direction + and the downwards direction −
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The ball's velocity is 65.1168 ft/sec when it hits the ground
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