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Grade: 12

                        

A long jumper leaves the ground at an angle of 60.0⁰ to the horizontal and at a velocity of 14.0m/s. Take g= 9.81m/s 2 . (a) How long does it take for him to reach maximum height? (b) What is the maximum height he reaches? (c) How far does he jump? (Assume the motion of his arm and legs is ignored).

3 years ago

Answers : (1)

Sachin
13 Points
							Formula for time of flight is 2usin#/g (where u is initial velocity and # is angle of progression) time to reach the max. height is usin#/g.14*sin60/9.81 = 1.235Formula is u^2sin^2#/2g14*14(sin60)^2/2g = 5.66Formula for range is u^2sin2#/g14*14sin120/9.81= 17.3014*14
						
3 years ago
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