Velocity of 1-dimensional projectile

[notReallyHere]

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

x_f=x_i+v_it+1/2at²
v_f=v_i+at

The Attempt at a Solution

For #1 I tried this:
x_i=202.5 x_f=0 a=-9.80 v_i=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)
V_f=0+(-9.8)(6.429)
V_f=-63.004

But that's wrong too since #1 is wrong. What did I do wrong in #1?

 

[Gib Z]

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!

 

[thomate1]

Your mistake in #1 is that you have used the wrong initial position (xi) and final position (xf) values. The initial position should be 85.0 m (the height at which the projectile is launched from) and the final position should be 0 m (the ground level where the projectile hits). So the correct equation would be:

xf = xi + vit + 1/2at^2

0 = 85 + 63t + 1/2(-9.8)t^2

-85 = 63t - 4.9t^2

4.9t^2 - 63t - 85 = 0

Solving this quadratic equation, we get two values for t: t = 6.429 s (which is the time it takes for the projectile to reach its maximum height) and t = 14.189 s (which is the time it takes for the projectile to hit the ground).

For #2, at the time of impact (t = 14.189 s), the velocity of the projectile can be found using the equation:

vf = vi + at

vf = 63 + (-9.8)(14.189)

vf = -63.004 m/s

Note that the negative sign indicates that the velocity is in the downward direction, which makes sense since the projectile is falling.