Question

While Standing on frictionless ice, you (mass 65.0 kg) toss a 4.50 kg rock with initial speed 12.0 m/s. The rock is 15.2 m from you when it lands. Take the x axis on the ice surface in the horizontal direction of the motion of the rock. Ignore the initial height of the toss. Part A At what angle did you toss it? Part B How fast are you moving? |v| =
how to do this question?

Answer 1

(A)

Applying momentum conservation to find speed of thrower after tossing

(65 + 4.50) x 0 = 4.50 x 12cos(theta) + 65 v

v = - 0.83cos(theta)

For horizontal Range

R = 15.2- vT

where T is the time of fi;ght. T = 2 v0 sin(theta) / g

15.2 - (0.83cos(theta) x 2 x 12 sin(theta) / 9.8) = v0^2 sin(2 theta) / g

15.2 - 1.02 sin(2 theta) = 12^2 sin(2 theta) / 9.81

14.69 sin(2 theta) + 1.02 sin(2 theta) - 15.2 = 0

sin ( 2 theta) = 0.967

2 theta = 75.3 deg

theta = 37.65 deg

(B) v = 0.83 cos(37.65) = 0.66 m/s