Question

# A diver leaves the end of a 4.0 m high diving board and strikes the water...

A diver 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. Ignoring the height of the diver (assume she is a particle), determine: a) her initial velocity and angle at which she jumped, b) her maximum height reached, and c) the velocity and angle with which she enters the water.

Let us consider:

u be the initial velocity of thee diver
x be the horizontal distance from the end of the diving board,
y0 be the initial height above water,
t be the time from take-off until entering the water,
g be the acceleration due to gravity,
a be the diver's angle to the horizontal at the time of take-off,
v be the diver's speed of entry into the water,
b be the angle of entry measured from the horizontal.

(a) 0 = y0 + ut sin(a) - gt^2 / 2 ...(1)
x = ut cos(a) ...(2)

From (1):
ut sin(a) = - y0 + gt^2 / 2 ...(3)

Sqaring (3) and adding to (2):
u^2 t^2 = x^2 + (gt^2 / 2- y0)^2
u^2 = [ x^2 + (gt^2 / 2 - y0)^2 ] / t^2
u = sqrt[ x^2 + (gt^2 / 2 - y0)^2 ] / t

y0 = 4.0 m
t = 1.3 s
x = 3.0 m
g = 9.81 m/s^2

Hence, u = sqrt[ 9.0 + (8.289 - 4.0)^2 ] / 1.3 = 4.02 m/s

From (2):
cos(a) = x / (ut) = 3/4.02*1.3 =0.574
a = 54.97 deg.

(b) As the vertical velocity at the highest point is 0:
0 = [ u sin(a) ]^2 - 2g(y - y0)
y = y0 + [ u sin(a) ]^2 / (2g)
= 4.0 + 0.552 = 4.552 m.

(c) v sin(b) = u sin(a) - gt ...(4)
v cos(b) = u cos(a) ...(5)

Squaring and adding (4) and (5):
v^2 = u^2 - 2ugt sin(a) + (gt)^2
v = [4.02^2 - 2*4.02*9.81*1.3*0.82 + (9.81*1.3)^2]^1/2

= [16.16 - 84.078 + 162.64]^1/2 = 9.73 m/s