Question

 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.​

Answer 1

Answer:

y = ut + $\frac{1}{2}$$at^{2}$ , y = 80t - $\frac{1}{2}$$32.2t^{2}$

for y = -200 ft,

-200 = 80t - $16.1t^{2}$ or $16.1t^{2}$ - 80t - 200 = 0

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

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

= 99.4 ft.

           

Explanation:

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

-Matt