Time remaining:
ball is thrown straight upward at an initial speed of v0 = 80 ft/s.h = −16t2 +

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If not possible, enter IMPOSSIBLE.)When does the ball initially reach a height of 64 ft?

) When does it reach a height of 118 ft? the greatest height reached by the ball?reach the highest POINT OF ITS PATH

When does the ball hit the ground 



Jan 26th, 2015

Since the ball feels the effects of gravity, use Constant Acceleration Equations:

HEIGHT of 118
x=vi*t+1/2*at^2
118 = 80t + 1/2*-9.8*t^2
118 = 80t - 4.9t^2
4.9t^2 - 80t + 118 = 0
Use quadratic formula to solve for t

GREATEST HEIGHT/HIGHEST POINT OF PATH, final velocity (vf = 0), acceleration is gravity (a = -9.8)
vf^2=vi^2+2ax
0=80^2+2(-9.8)x
19.6x=6400
x = 326.5

HITS GROUND at twice the time it takes to reach highest point
vf=vi+at
At highest point, vf=0
0 = 80 + -9.8*t
9.8t = 80
t = 8.16
2t = 16.32

Jan 26th, 2015

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Jan 26th, 2015
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