Question

#10

Problem 3.18 Constants A diver running 2.5 m/s dives out horizontally from the edge of a vertical cliff and 2 5 s later reaches the water below How high was the citf? How tar trom its base did the diver hit the wate? Express your answer using two significant figures Provide Feedback

Answer 1

Initial velocity of diver $u_x=2.5\,m/s$ (only in horizontal direction), $u_y=0\,m/s$ (no vertical velocity)

Time taken for the diver to reach water surface $t=2.5\,s$

Part A)

Using $y=u_yt+\frac{1}{2}a_yt^2$

Vertical distance $y=-h$, $u_y=0\,m/s$, Vertical acceleration $a_y=-g$

Height of the cliff $-h=u_yt-\frac{1}{2}gt^2$ (all downward vectors are taken negative)

$-h=0*2.5-\frac{1}{2}*9.81*2.5^2=-30.7\,m$

$h=30.7\,m\approx 31\,m$

Part B)

Using$x=u_xt+\frac{1}{2}a_xt^2$

Horizontal distance $x=d$, $u_x=2.5\,m/s$  , $a_x=0$

Distance of base of cliff to water where the diver reached is $d=u_xt=2.5*2.5=6.25\,m\approx 6.2\,m$