2. Samuel leaves the end of a 4.0-m-high diving board and strikes the water 1.3 s later, 3.0 m beyond the end of the board. Considering the
diver as a particle, determine:
(a) His initial velocity,
(b) The maximum height reached;
(c) The velocity Vf
with which he enters the water.

a) v = 4.9 m/s at 62° above the horizontal

b) h = 0.94 m above the board or 4.94 m above the water

c) v = 10 m/s at 77° below the horizontal

Explanation:

Assume UP and FOREWARD are positive and the diving board is the origin.

Ignore air resistance.

In the vertical

-4 = 0 + vy(1.3) + ½(-9.8)(1.3²)

vy = 4.281

in the horizontal

3.0 = 0 + vx(1.3) + ½(0)(1.3²)

vx = 2.307...

v = √(4.281² + 2.307²) = 4.863...

tanθ = 4.281 / 2.307

θ = 61.680...

v² = u² + 2as

s = (v² - u²) / 2a

h = (0² - 4.281²) / 2(9.8) = 0.9350...

v² = 4.9² + 2(9.8)(4)

v = 10.1019

cosθ = 2.307/10,1019

θ = 76.798

not too sure thinking all accelerations are pointed down