Physics
posted by Jamie
A projectile is launched from ground level to the top of a cliff which is 195 m away and 155 m high. If the projectile lands on top of the cliff 9.2 s after it is fired, find the initial velocity of the projectile (magnitude and direction). Neglect air resistance.

Henry
d = Vo*t + 0.5at^2,
We have one Eq and 2 unknowns: Vo, & a.
Vo = at, a = Vo / t,
In the 1st Eq, substitute Vo / t for a:
d = Vo*t + 0.5(Vo/t)*t^2,
Simplify:
d = Vo*t + 0.5Vo*t,
195 = 9.2Vo + 0.5*9.2Vo,
195 = 9.2Vo + 4.6Vo,
195 = 13.8Vo,
Vo = 195 / 13.8 = 14.13 m/s = Initial
velocity.
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