A ball is thrown vertically up with a velocity of 80 ft/sec at the edge of a 200-ft cliff. Calculate the height h to which the ball rises and the total time t after release for the ball to reach the bottom of the cliff. Neglect air resistance and take the downward acceleration to be 32.2 ft/sec.​

Answers

Answer 1

Answer:

y = ut + [tex]\frac{1}{2}[/tex][tex]at^{2}[/tex] , y = 80t - [tex]\frac{1}{2}[/tex][tex]32.2t^{2}[/tex]

for y = -200 ft,

-200 = 80t - [tex]16.1t^{2}[/tex] or [tex]16.1t^{2}[/tex] - 80t - 200 = 0

t = 80 ±  [tex]\frac{\sqrt{80^{2} + 4(16.1)(200) }}{2(16.1)}[/tex] = 6.80 sec. ( or - 1.83 s)

For y = 0,  [tex]u^{2}[/tex] = [tex]u_{o}^{2}[/tex] + 2ay, y = h = [tex]\frac{0-80^{2} }{-2(32.2)}[/tex]

= 99.4 ft.

           

Explanation:

Hope this helped, please message me if I made a mistake!

-Matt


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Answers

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