A ball is thrown into the air with an initial velocity of 64 feet per second from a height of 336 feet. After how many seconds will the ball return to the ground and have a height of zero again
Answer
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Answer:
7 seconds
Step-by-step explanation:
Ballistic motion is usually modeled in the vertical direction in US customary units by the equation h(t) = -16t^2 +v0·t +h0, where v0 and h0 are the initial velocity and height, and h(t) is the height as a function of time in seconds. For the given initial conditions, the equation of vertical motion will be ...
h(t) = -16t^2 +64t +336
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This question asks you for the value of t for which h(t) = 0. We can solve that equation by factoring.
0 = -16t^2 +64t +336
0 = t^2 -4t -21 . . . . . . . . divide by -16
0 = (t -7)(t +3) . . . . . . . factor the quadratic
t = 7 . . . . . . the positive value of t that makes the equation true
The ball will return to the ground after 7 seconds.
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Additional comment
A graph of the function reveals the ball reaches a maximum height of 400 feet after 2 seconds.
In metric units, the equation is h(t) = -4.9t^2 +v0·t +h0, where distances are in meters instead of feet. Time is still in seconds.