A diver leaves a 3-m board on a trajectory that takes her 2.7m above the board and then into the water 2.9m horizontally from the end of the board.At what speed did she leave the board?At what angle did she leave the board?

Answer 1

h = 0.5a*t^2 = 3+2.7 = 5.7 m.
4.9t^2 = 5.7
t^2 = 1.16
Tf = 1.08 s. = Fall time.

Dx = Xo*Tf = 2.9 m
Xo * 1.08 = 2.9
Xo = 2.69 m/s = Hor. component of initial velocity.

Y^2 = Yo^2 + 2g*h = 0
Yo^2 = -2g*h = -2*(-9.8)*2.7 = 52.92
Yo = 7.27 m/s = Ver. component of initial velocity.

Tan A = Yo/Xo = 7.27/2.69 = 2.70432
A = 69.7o

Vo = Yo/sin A = 7.27/sin69.7 = 7.75 m/s

Answer 2

Correction:

h = 0.5g*t^2 = 2.7
4.9t^2 = 2.7
t^2 = 0.551
Tf1 = 0.74 s = Fall time to board.

Tr = Tf1 = 0.74 s = Rise time.

h = 0.5g*t^2 = 3 + 2.7 = 5.7
4.9t^2 = 5.7
t^2 = 1.16
Tf2 = 1.08 s = Fall time from max ht. to
the water.

Dx = Xo*(Tr+Tf2) = 2.9 m.
Xo = 2.9/(0.74+1.08) = 1.59 m/s = Hor.
component of initial velocity.

Y^2 = Yo^2 + 2g*h = 0
Yo^2 = -2g*h = -2*(-9.8)2.7 = 52.92
Yo = 7.27 m/s = Ver. component of initial velocity.

Tan A = Yo/Xo = 7.27/1.59 = 4.56255
A = 77.6o

Vo = Yo/sin A = 7.27/sin 77.6 = 7.44 m/s
[77.6o] = Speed and angle at which she
left the board.