When standing on the edge of a cliff that is 125 meters high, you through a phone at an angle of 65.0 degrees above horizontal at a speed of 12.32 meters per second.

Question

-How long is it in the air?

-How far does it go in the horizontal direction?

-What maximum height does it reach?

Answer 1

First calculate initial velocity components: V_xo = 12.32 cos 65.0 = 5.207 m/s; v_yo = 12.32 sin 65.0 = 11.17 m/s. We also take vertical acceleration a to be -9.81 m/s².

a) Δy = v_yo t + 1/2 a t² where Δy = y_f - y_i = -125m = 11.17 t - 4.905 t², 4.905 t² - 11.17 t - 125 = 0, t = {11.17 ± √[11.17² - 4 (4.905) (-125)] } /9.81 --> total t = {11.17 + 50.77} /9.81 = 6.31 s.

b) The horizontal motion assumes zero acceleration so v_x is constant. Δx = v_x t = (5.207)(6.31) = 32.9 m.

c) At maximum, v_yf = 0, v_yf² = v_yo² + 2a Δy = 0 = (11.17)² - (19.62) Δy, Δy = 6.35 m so the height above ground level is 131 m.