Velocity of 1-dimensional projectile

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The projectile is launched upward at +63.0 m/s from a height of 85.0 m, reaching a maximum height of 202.5 m before falling. To find the time to hit the ground, the total time includes both the ascent and descent phases. The initial approach incorrectly calculated time by only considering the descent from the maximum height. The correct method requires calculating the time to reach the peak and then the time to fall from that peak to the ground. Accurate calculations are essential for determining the velocity at impact, which depends on the correct time derived from both phases of 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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