[high dive) above a pool of water. according to the announcer, the divers enter the water at a speed of 56 mi/h (25 m/s). (air resistance may be ignored in this problem.) ve] visitors at an amusement park watch divers step off a platform 70 ft (a) is the announcer correct in this claim? explain. (b) is it possible for a diver to leap directly upward off the board so that, missing the platform on the way down, the diver enters the water at 25 m/s? if so, what initial upward speed is required? is the required initial speed physically attainable? how do you know?

Distance^2= yellow line ^2 (see picture) = 30^2 + 49^2so distance = squareroot of (300^2 +49^2)i dont know if this is how to solve it, but i cnt think of another way

Dis the

The answer which is not a detail carson used to describe the pool in the cave at low tide is b.  hundreds of periwinkles browse the branches and roots of the trees. 

W= fd
where f is an applied force and d is the distance over which it is applied. when a parachuting skydiver is at constant velocity, he is experiencing "terminal velocity", the point at which his acceleration down is canceled out by air resistance. that means that the resisting force of the air is equal to his weight, but in the opposite direction. with this, we can solve for the work done by the force.
w = fd
w = (92.0)(9.81 m/s^2) (325)
w= 2.93 x 10^5 j
the value for work in this case is positive because it is pushing up on the skydiver, who has a downward force of gravity.
therefore the answer is 2.93 x 10^5 j