Question

6. You throw a ball out of a window at 4 m/s at a downward 30° angle. As a result, it takes 1.5 seconds for the ball to hit the ground. 30 A. How far does this ball travel horizontally before hitting the ground? (5.2 m) B. How high is this window from the ground? C. What is the velocity of this ball just before it hits the ground? Report your answer using components, or calculate the magnitude and direction.

Answer 1

(a) use S = ut + 1/2 at^2 for x-axis x = 4 cos 30 degree times 1.5 x = 5.2 m (b) Let the height of window is h S = -h, u_y = -4 sin 30 degree, a_y = -g, t = 1.5 m use II equation of motion S = u_y t + 1/2 a_y t^2 -h = -4 sin 30 degree times 1.5 - 1/2 times 9.8 times 1.5^2 -h = -3 - 11.025 h = 14.025 m (c) Let it strikes the ground with speed v v_x = u_x = 4 cos 30 degree = 3.464 m/s v_y = u_y + a_y t = -4 sin 30 degree - 9.8 times 1.5 v_y = -2 - 14.7 = -16.7 m/s v = squareroot v_x^2 + v_y^2 = squareroot (3.464)62 + (-16.7)^2 v = 17.06 m/s tan theta = v_y/v_x = -16.7/3.464 = -4.82 theta = 78.28 degree (from - x-axis) Angle from + axis, angle is 180 + 78.28 = 258.28 degree