Question

A projectile is launched straight up at 54.5 m/s from a height of 88 m, at the edge of a sheer cliff. The projectile falls, just missing the cliff and hitting the ground below.

(a) Find the maximum height of the projectile above the point of firing.
151.5
I calculated and found the maximum height just struggling for the other parts

(b) Find the time it takes to hit the ground at the base of the cliff.
s

(c) Find its velocity at impact.
m/s

Answer 1

Hi = 88
Vi = +54.5
a = -9.81

v = Vi - 9.81 t
v = 0 at top
so at top t = 54.5 / 9.81 = 5.56 seconds
then h = Hi + Vi t - 4.9 t^2
= 88 + 54.5 (5.56) - 4.9 (5.56)^2 = 88 + 303 -151 = 240 meters which is 240-88 = 152 meters above firing point

when does h = 0 ?
0 = 88 + 54.5 t - 4.9 t^2
solve quadratic for t
https://www.mathsisfun.com/quadratic-equation-solver.html
roots are t = -1.43 or t = +12.5
the negative root is when it would have been on the ground before the firing
use t = 12.5 seconds total in air

Then again v = Vi - 9.81 t
= 54.5 - 9.81 (12.5)

Answer 2

a. Vo^2 + 2g*h = V^2.
54.5^2 + (-19.6)h = 0,
h = 151.5 m.

b. ho + h = 88 + 151.5 = 239.5 m. above gnd.
0.5*g*Tf^2 = 239.5.
0.5*9.8*Tf^2 = 239.5,
Tf = 7 s. = Fall time.

Vo + g*Tr = V.
54.5 + (-9.8)Tr = 0,
Tr = 5.6 s. = Rise time.
T = Tr + Tf = 7+ 5.6 = 12.6 s. = Time to reach gnd.

c. V^2 = Vo^2 + 2g*h = 0 + 19.6 *239.5 = 4694.
V = 68.5 m/s.