A man jumps horizontally from the top of a building that is 20.0m high, and hopes to reach a swimming pool that is at the bottom of the building, 10.0m horizontally from the edge of the building. If he is to reach the pool, what must his jumping speed be?

5 m/s

Explanation:

Given that the hight of the building, h= 20.0 m

The horizontal distance of the swimming pool from the building, d=10 m

As the man jumps horizontally, so the vertical velocity at the time of jumping is 0.

Let s be the initial horizontal velocity of the man at the time of jumping.

As the gravitational force always acts in the vertical direction, so, it does not change the horizontal speed.

So, the horizontal velocity, s, remains constant throughout the motion.

If t be the time of flight, then

Now, applying the equation of motion in the vertical direction,

where u is the initial velocity, t is the time of flight, a is the acceleration, s is the displacement.

Here, putting s= 20 m, u=0, in the equation of motion, we have

seconds (approx)

Now, put t=2 in the equation ( we have

m/s

Hence, the man must jump with 5 m/s horizontally to reach the pool.

the answer is a. the yacht traveled from the dock north to the 200-meter marker in the bay in less than 5 minutes, giving the passengers several more hours to fish.

im doing peak rn so sry if its wrong