Question

A projectile is launched from the edge of a 50 m high vertical cliff at an angle of 30° above horizontal. The projectile hits the ground at a distance of 150 m from the bottom of the cliff. Find the initial and final velocities of the projectile (magnitude and direction) as well as the total displacement performed by the projectile.

Answer 1

In horizontal it went 150 m with ucos30 .

so t = 150 / ucos30

ut = 173.20 ..... (i)

In vertical, Using H = ut + at^2 /2

-50 = (usin30) t - 9.81t^2 /2

-50 = 0.5ut - 4.905t^2 where ut = 173.20

putting in

-50 = 0.5*173.20 - 4.905t^2

t =5.28 sec .

Initial velocity u = 173.20 /t = 173.20 / 5.28 = 32.82 m/s

angle = 30 degrees above horizontal .

for final :

vx = ux = 32.82cos30 = 28.42 m/s

vy = uy + at = 32.82sin30 - 9.81*5.28 = - 35.39 m/s

magnitude = sqrt(vx^2 + vy^2) =45.39 m/s

angle = tan^-1(vy/uy ) = 51.23 degrees below the horizontal.

Total displacement = sqrt(x^2 + y^2) =sqrt(50^2 + 150^2) = 158.11 m