Question

A cliff diver jumps off a 45m cliff and lands in the water 3.20 seconds later, a distance of 1.75m from the cliffs edge. What was the magnitude and direction of his starting velocity? What was the diver's speed upon entering the water?

Answer 1

here,

the height of cliff , h = 45 m

time taken , t = 3.2 s

let the magnitude and direction of initial velocity be u and theta

for vertical direction

-h = u * sin(theta) * t - 0.5 * g * t^2

- 45 = u * sin(theta) * 3.2 - 0.5 * 9.81 * 3.2^2 ...(1)

and

for horizontal direction

x = u * cos(theta) * t

1.75 = u * cos(theta) * 3.2 ....(2)

from (1) and (2)

u = 1.72 m/s

theta = 71.5 degree

the initial velocity is 1.72 m/s and 71.5 degree above the horizontal

let the speed be v before hitting the water

using conservation of mechanical energy

0.5 * m * v^2 = 0.5 * m *u^2 + m * g * h

0.5 * v^2 = 0.5 * 1.72^2 + 9.81 * 45

v = 29.8 m/s

the final speed is 29.8 m/s