A projectile is shot from the edge of a cliff 120 m above ground level with an initial speed of 60 m/s at an angle of 30.0o with the horizontal, as shown below. (a) How long does it take for the projectile to hit the ground? (b) How far does the projectile land from the base of the cliff? (c) What are the horizontal and the vertical components of its velocity when it hits the ground? (d) What are the magnitude of the velocity, and the angle made by the velocity vector with the horizontal. (e) What’s the maximum height above the cliff reached by the projectile. (f) How long does it take for the projectile to reach the maximum height?

Answer 1

Vertical problem:
Vi = 60 sin 30 = 30 m/s upwards
Hi = 120 meters high
a = g = -9.81 m/s^2
so
v = Vi - g t = 30 - 9.81 t
h = Hi + Vi t - (1/2) a t^2 = 120 + 30 t - 4.9 t^2
when does h = 0 (ground) ?
4.9 t^2 - 30 t - 120 = 0
solve quadratic
(a) t = +8.88 seconds or -2.76 seconds
use the +8.88. the negative answer was on the way up if we had shot it before t = 0
(b) u = horizontal speed = 60 cos 30 = 52m/s forever and ever while above ground
so horizontal distance = u t = 52 *8.88 = 461 meters
(c) v = Vi - g t = 30 - 9.81 * 8.88 = -57.1 m/s
and of course u = 52 m/s
(d) speed = sqrt (52^2+57.1^2)
tangent of angle below horizontal = 57.1 / 52

(e) and (f) back to part (a)
v = Vi - g t = 30 - 9.81 t
when is v = 0 (at the top)
t = 30/9.81 = 3.06 seconds to top
h above cliff = Vi t - 4.9 t^2 = 30(3.06) - 4.9(3.06)^2
= 91.8 - 45.9 = 45.9 meters above the cliff