Question

A large rock is propelled off the edge of a cliff at an initial velocity of 14 m/s and an angle of 45

Answer 1

You need component velocities of the launch velocity and angle.
Horizontal component = (cos 45) x 14 = 31.82m/sec.
vertical component = (sin 45) x 14 = 31.82m/sec^2.
(Note both are identical, as the angle involved is 45 deg. You could consider the 14m/sec as the diagonal of a square, and the length of the sides is (1/2 (sqrt.2)) x 14. This saves trig. calcs.).
At 31.82m/sec vertical velocity, the rock will rise above the cliff by (v^2/2g), = 51.66m.
It will take sqrt. (2h/g) = 3.25s. to reach max. height.
It then falls (130 + 51.66) = 181.66m. to the ground. This takes sqrt. (2h/g) = 4.083s. to happen.
a) Total time in air = (4.083 + 3.25) = 7.333s.
b) In 7.333s., using the above horizontal velocity of 31.82m/sec., it will have travelled (31.82 x 7.333) = 233.34m. from the cliff base.

Vertical velocity component at ground = sqrt. (2gh), = sqrt.(19.6 x 181.66), = 59.67m/sec.
Horizontal velocity = 31.82m/sec. Use Pythagoras to find impact velocity.
c) Sqrt.(59.67^2 + 31.82^2) = 67.62m/sec., and atn. (31.82/59.62) = 28.1 deg. from vertical.
Edit, I used 9.8 as g.