SOLUTION: I do not understand this:
A ball is thrown vertically upward with an initial velocity of 160 feet per second. The distance in feet of the ball from the ground after 1 second is
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Question 134996: I do not understand this:
A ball is thrown vertically upward with an initial velocity of 160 feet per second. The distance in feet of the ball from the ground after 1 second is s=160t-16t^2. For what intervals of time is the ball less than 336 above the ground(after it is tossed until it returns to the ground)
Good luck and thanks
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
A ball is thrown vertically upward with an initial velocity of 160 feet per second. The distance in feet of the ball from the ground after 1 second is s=160t-16t^2. For what intervals of time is the ball less than 336 above the ground(after it is tossed until it returns to the ground)
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Find out when the ball is at 336 ft.
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-16t^2+160t = 336
Divide thru by -16 to get:
t^2 - 10t -21 = 0
(t-7)(t-3) = 0
t = 3 seconds and t= 7 seconds.
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The ball is below 336 when 0<=t<3 seconds on the way up.
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The ball is below 336 when 7
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Comment: It hits the ground when t = 10 seconds the time
up and the time down are symmetric.
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Cheers,
Stan H.
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