Question

(3) A professional diver jumps (up) off of a cliff that is 18 m high. Draw a sketch of the cliff, defining your origin and final position. (Careful with negative and positive signs.) Unlike the WB ssignment, assume the diver jumps up first and has an initial vertical velocity is 2.5 m/s. (Ignore air resistance.) (a) How long does it take the diver to hit the water? (b) How high from the top of the cliff does the diver reach before starting to fall down? (c) What's the diver's velocity on impact with the water? (Careful with negative and positive signs.) (d) Compare this velocity with the answer you got for WB 3. Is it more or less? About the same? Why? (You have to write a brief explanation. Please don't

Answer 1

We will use equatins of motion in straigh line with constant acceleration to solve this problem.

u=initial velocity, v =final velocity ,a=-9.8 m/s² , s=displacement

at highest point velocity=0

We will use equatins of motion in straigh line with constant acceleration to solve this problem. u = initial velocity, v = final velocity, a = minus 9.8 m/s^2, s = displacement at highest point velocity = 0 (a) s = ut + 1/2 at^2 = minus 18 = (2 minus 5) t + 1/2 (minus 9.8) t^2 = 4.9 t^2 minus 2.5 t minus 18 = 0 t = 2.19 s (b) v^2 = u^2 + 2 as (0)^2 = (2.5)^2 + (2) (minus 9.8) s s = (2.5)^2/19.6 s = 0.32 m (c) v = u + at v = (2.5) + (minus 9.8) (2.19) v = minus 18.95 m/s