A high diver leaves the end of a 6.0 m high diving board and strikes the water 1.4 s later, 4.0 mbeyond the end of the board. Considering the diver as a particle,
Determine the magnitude of her initial velocity, v⃗ 0.
Express your answer to two significant figures and include the appropriate units.
Determine the angle of her initial velocity, v⃗ 0.
Express your answer to two significant figures and include the appropriate units.
Determine the maximum height reached.
Express your answer to two significant figures and include the appropriate units.
Determine the magnitude of the velocity v⃗ f with which she enters the water.
Express your answer to two significant figures and include the appropriate units.
Determine the angle of the velocity v⃗ f with which she enters the water.
Express your answer to two significant figures and include the appropriate units.
in horizontal,
(v0 cos(theta)) t = 4
v0 cos(theta) = 4/1.4 = 2.857 ..... (i)
in vertical,
yf - yi = v0y t + ay t^2 /2
0 - 6 = (v0 sin(theta))(1.4) + (-9.8 x 1.4^2 / 2)
v0 sin(theta) = 3.604 .... (ii)
(ii)/(i) => tan(theta) = 3.604 / 2.857
theta = 51.6 deg ....Ans(Angle of initial velocity)
v0 = 4.60 m/s ......Ans (magnitude of initial velocity )
Hmax = 3.604^2 / (2 x 9.8) = 0.66 m from board
and 0.66 + 6 = 6.66 m from water level
vx = v0x = 2.857 m/s
vy = (3.604) + (-9.8x 1.4) = -10.116 m/s
magnitude of vf= sqrt(vx^2 + vy^2) = 10.5 m/s ....Ans
angle = tan^-1(-10.116/2.857) = -74.2 deg
Or 74.2 below the horizontal
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