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Olympic diver Matthew Mitcham springs upward from a diving board that is 3.20 m above the water. He enters the water at a 73.0...Question
Olympic diver Matthew Mitcham springs upward from a diving board that is 3.00 m above the water. He enters the water at a 79.0 degree angle with respect to the water surface, at a speed of 9.01 m/s. Determine the magnitude of his initial velocity. Determine the direction of his initial velocity, in terms of degrees relative to horizontal. Determine his maximum height above the water.
Answers
whip
First we must break it up into its individual x and y directions.
velocity in y direction =9.01*cos(11)= 8.8445 m/s downwards
velocity in x direciton =9.01*cos(79)=
1.71918 m/s to the right.
Given Vfy(velocity final y direction) we can find max height by using V=Vi+At acceleration being -9.81m/s^2.
using that formula we get 3.9869 meters as the max height reached.
As the diver starts 3 meters above the water 3.9869m-3m=.9869 m we can use this to find initial velocity in y direction using equation (vf)^2-(vi)^2=2a(xf-xi)
plugging this in given vfy = 0 at the top of the curve we get Viy= 4.4003 m/s
Because there is no horizontal acceleration Vix=1.71918 m/s (found at beginning). To get the magnitude of this velocity we use the Pythagorean thrm. 4.4003^2+1.7198^2=(4.724)^2
initial magnitude=4.724 m/s
Finally to determine direction we use inverse cosine of Vx/Vmagnitude to get a final answer of 68 degrees with respect to the horizontal.
equations used:
v²-v₀²=2a(x-x₀) v = v₀ + at
a^2+b^2=c^2 and inverse cosine
velocity in y direction =9.01*cos(11)= 8.8445 m/s downwards
velocity in x direciton =9.01*cos(79)=
1.71918 m/s to the right.
Given Vfy(velocity final y direction) we can find max height by using V=Vi+At acceleration being -9.81m/s^2.
using that formula we get 3.9869 meters as the max height reached.
As the diver starts 3 meters above the water 3.9869m-3m=.9869 m we can use this to find initial velocity in y direction using equation (vf)^2-(vi)^2=2a(xf-xi)
plugging this in given vfy = 0 at the top of the curve we get Viy= 4.4003 m/s
Because there is no horizontal acceleration Vix=1.71918 m/s (found at beginning). To get the magnitude of this velocity we use the Pythagorean thrm. 4.4003^2+1.7198^2=(4.724)^2
initial magnitude=4.724 m/s
Finally to determine direction we use inverse cosine of Vx/Vmagnitude to get a final answer of 68 degrees with respect to the horizontal.
equations used:
v²-v₀²=2a(x-x₀) v = v₀ + at
a^2+b^2=c^2 and inverse cosine