Question

A projectile fired from the edge of a 150 m high cliff with an initial velocity of 180 m/s at an angle of elevation of 30 deg with the horizontal.
i. The greatest elevation above the ground reached by the projectile.
ii. Horizontal distance from the gun to the point, where the projectile strikes the ground.

Answer 1

vertical problem
Hi = 150 m
Vi = 180 sin 30 = 90 m/s
v = Vi - g t = 90 - 9.81 t
v = 0 at top so t = 9.17 seconds to top
h = Hi + Vi t - (g/2) t^2

h at top = 150 + 90(9.17) - 4.9(9.17^2)
= 150 + 825 - 412 = 563 meters above ground at top after 9.17 s

now how long before ground?
falls from 563 meters high
v = - g T
0 = 563 - 4.9 T^2
T = 10.7 seconds falling
total time = t + t = 9.17 + 10.7 = 19.9 seconds
horizontal speed = 180 cos 30 = 156 m/s
so 156 m/s for 19.9 seconds ---> 3102 meters horizontal to target

Answer 2

KEEP UP

Answer 3

Given: 180m/s - initial velocity, 150m- height of cliff, 30° angle
Unknown: H- greatest elevation, R- horizontal distance until it strikes the ground
H=150m + (Vosin30°)² /2g
H=150 + (180m/s sin30°)² /2(9.81m/s²)
=562.844meters

R=(Vo² sin2(30°))/g
R=((180m/s)²sin(60°))/(9.71m/s²)
R=2860.267 meters