Question

A motorcycle stunt rider rides off the edge of a cliff. Just at the edge his velocity is horizontal, with a magnitude of 9.0 m/s. After 0.50 seconds find the motorcycle’s: a) position, b) distance from the edge of the cliff, and c) the resultant velocity.

Answer 1

a) initial speed is u.

then horizontal component u_x = 9 m/s

vertical component u_y = 0 ( initial velocity is in horizontal direction)

using r = ut + at^2 /2

for t = 0.50 s

x = (9*0.50) - (0 * 0.50^2 /2) = 4.5 m

y = (0*0.50) - (9.81 * 0.50^2 /2 ) = - 1.23 m

so 4.5m away horizontallya and 1.23m vertically down.

b) distance = sqrt(x^2 + y^2) = sqrt(4.5^2 + 1.23^2) = 4.66m

c) v = u + at

so v_x = 9 + 0 = 9 m/s

v_y = 0 + (-9.81*0.50) = 4.90 m/s

magnitude = sqrt(v_x^2 + v_y^2) = 10.25 m/s

direction = tan^-1(4.90/9) =28.59 deg below horizontal.