question 3 and 4 Plz 1.00 m diving board 10.0 m A diver with a mass of 69.7 kg d
ID: 1877013 • Letter: Q
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question 3 and 4 Plz
1.00 m diving board 10.0 m A diver with a mass of 69.7 kg dives off a diving board 10.0 m above the pool. When the diver launches himself he makes an angle of e 82.8 with the horizontal. For this question assume you can treat the diver as a particle located at the center of mass of the diver. The diver's center of mass is 1.00 m above the bottom of his feet. During the dive the diver's center of mass reaches a maximum height of 1.03 m above its initial height. The figure above is not to scale. How long after take off does the diver reach maximum height? 0.458 Check what is the magnitude of the initial velocity of the diver? 4.5288 m/'s How far horizontally from the take off point does the diver hit the water? CheckExplanation / Answer
3.
Time taken by diver to hit the water will be given by:
S = Uy*t + 0.5*a*t^2
S = -(10 m + 1 m) = -11 m (distance traveled by center of mass of diver)
Uy = U*sin A = 4.5288*sin 82.8 deg
Uy = 4.4931 m/sec
a = -g = -9.81 m/sec^2
Using these values:
-11 = 4.4931*t - 0.5*9.81*t^2
4.905*t^2 - 4.4931*t - 11 = 0
Solving above quadratic equation
t = 2.024 sec (take positive root of above equation)
Now We know that range in projectile motion is given by:
Range = Ux*T
T = time period of motion = 2.024 sec
Ux = U*cos A = 4.5288*cos 82.8 deg
Range = 2.024*4.5288*cos 82.8 deg = 1.1488 m
Part D.
Using energy conservation on both points
KEi + PEi = KEf + PEf
KEi = initial kinetic energy of diver when he jumped = 0.5*m*U^2
PEi = initial potential energy of diver when he was on board = m*g*h
KEf = final KE of diver = ?
PEf = 0, Assuming surface of water as reference point
So,
0.5*m*U^2 + m*g*h = KEf
KEf = 0.5*69.7*4.5288^2 + 69.7*9.81*11
KEf = 8236.1 J
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