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A diver jumps off a cliff of height h at an angle θ = 430  (see figure)....

A diver jumps off a cliff of height h at an angle θ = 430  (see figure). He reaches a maximum height of = 15 m above the top of the cliff before falling to the water below. He hits the water x=295 m from the base of the cliff. Determine  the initial speed of the diver,  . Take  g=10m/s2 . Round your answer to one decimal place

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Answer #1

Diver is like a projectile shot at 43° which reaches a maximum height of 15 m above the ground.

Let the initial velocity of the diver be u.

Vertical component of the velocity is .

Diver is under constant acceleration due to gravity, force of gravity is being applied against his vertical motion. Therefore, maximum height reached is given by

When the diver reaches maximum height above the cliff, his vertical velocity is zero, .

Maximum height of the diver is

, where u is the initial speed.

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