Question

A daring 510-N swimmer dives off a cliff with a running horizontal leap. What must her minimum speed be just as she leaves the top of the cliff so that she will miss the ledge at the bottom, which is 1.75 m wide and 9.00 m below the top of the cliff?

Answer 1

Answer:

$v_x = 1.26 m/s$

Explanation:

given,

weight of swimmer = 510 N

length of ledge, L = 1.75 m

vertical height of the cliff, h =  9 m

speed of the swimmer = ?

horizontal velocity  of the swimmer should be that much it can cross the wedge.

distance = speed x time

d = v_x × t

1.75 = v_x × t ........(1)

now,time taken by the swimmer to cover 9 m

initial vertical velocity of the swimmer is zero.

using equation of motion for time calculation

$s = ut +\dfrac{1}{2}gt^2$

$9= 0+\dfrac{1}{2}\times 9.8\times t^2$

  t² = 1.938

  t = 1.39 s

same time will be taken to cover horizontal distance.

now, from equation 1

1.75 = v_x × 1.39

$v_x = 1.26 m/s$

horizontal speed of the swimmer is equal to 1.26 m/s