Velocity of 1-dimensional projectile

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    Projectile Velocity
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SUMMARY

The discussion focuses on the calculations involved in determining the time it takes for a projectile, launched at +63.0 m/s from a height of 85.0 m, to hit the ground below a cliff. The maximum height reached by the projectile is 202.5 m. The initial attempts to calculate the time of flight and impact velocity were incorrect due to neglecting the ascent time. The correct approach requires considering both the ascent and descent phases of the projectile's motion.

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Homework Statement


A projectile is launched straight upward at +63.0 m/s from a height of 85.0 m, at the edge of a sheer cliff. The projectile falls, just missing the cliff and hitting the ground below.

Maximum height = 202.5m

1) Find the time it takes to hit the ground at the base of the cliff.
2) Find its velocity at impact

Homework Equations


xf=xi+vit+1/2at2
vf=vi+at

The Attempt at a Solution


For #1 I tried this:
xi=202.5 xf=0 a=-9.80 vi=0
0=202.5+0+1/2(-9.8)t2
-202.5=-4.9t2
41.327=t2
t=6.429

But that's not the right answer

For #2 I tried this: (It depends on the answer to #1)
Vf=0+(-9.8)(6.429)
Vf=-63.004

But that's wrong too since #1 is wrong. What did I do wrong in #1?
 
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By putting x_i = 202.5, you are only solving for the time it takes to fall from the top of its motion, down to the ground. You have to account for the time it took to get up there!
 

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