A diver initially moving horizontally with speed v dives dives offthe edge of a vertical cliff and lands in the water a distance dfrom the base of the cliff. How far from the base of the cliffwould the diver have landed if the diver initially had been movinghorizontally with speed 2v?

A. d

B. √ (2d)

C. 2d

D. 4d

E. It cannot be determined unless the hieght of the cliff isknown

Option C is correct

Explanation:

"The time is determined by the vertical distance. The formula is sqr (2d/a) = t. There still is no acceleration in the horizontal direction."

For the first drive

d = d

t = sqr (2d/a)

r = v

For the Second drive

d = ?

t = sqr (2d/a)

r = 2v

Since the times are same, equate the results.

t = d/v = d1/2v

v*d1 = 2v*d The v's cancel because they are related.

d1 = 2d.

You go twice as far as you did before.

Option C is correct