Answer:
(a) 13.795 m/s.
(b) -3140.28 N.
Explanation:
(a) Using newton's equation of motion,
v² = u² + 2gs.......................... Equation 1
Where v = final velocity, u = initial velocity, s = height of the tower, g = acceleration due to gravity.
Given: s = 9.7 m, u = 0 m/s ( jump from a height), g = 9.81 m/s².
Substitute into equation 1
v² = 0² + 2×9.81×9.7
v² = 190.314
v = √(190.314)
v = 13.795 m/s.
Hence the velocity of the driver when he hits the water = 13.795 m/s.
(b)
F = ma.................... Equation 2
Where F = force exerted on the diver, m = mass of the diver, a = acceleration of the diver below the water surface.
Also using
v² = u² + 2as ............ Equation 3
Note: At the point when the diver enters the water, u = 13.795 m/s, and at the point when the diver comes to a complete stop, v = 0 m/s
Given: s = 2.0 m, u = 13.795 m/s, v = 0 m/s
Substitute into equation 3
0² = 13.795²+2(2a)
0 = 190.30203 + 4a
-4a = 190.30203
a = 190.30203/-4
a = -47.58 m/s²
Also given: m = 66 kg,
Substitute into equation 3
F = (-47.58)(66)
F = -3140.28
Note: The Force is negative because it act against the motion of the diver.
Hence the net force exerted on the diver while in the water = -3140.28 N.
