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A projectile is fired at time t = 0.0s, from point 0 at the edge of a cliff, with initial velocity components Vox = 80 m/s and and Voy = 800 m/s. The projectile rises, then falls into the sea at point P. The time of flight of the projectile is 200.0 s

What is the height of the cliff?
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Voy = 800m/s

time after which it will reach the level from which it was thrown is give by : s = ut+1/2at^2

here u = 800m/s

s=0

a = -9.8m/s2

thus t = 163.3s

velocity at that time = u+at = 200 - 9.8*163.3 = -800 (i.e is downward direction)

that will be treated as u for this next section.

time left in flight = 200- 163.3 = 36.7 s

thus heigh = H = ut+1/2at^2 = 35959.8 m

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