InterviewSolution
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InterviewSolution
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A ball is projected vertically upward with speed of 50 m/s. Find a. the maximum height, b. the time to reach themaximum height, c. the speed at half the maximum height. Take g=`10 m/s^2`. |
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Answer» Correct Answer - B::C Given, `u=50 m/s g=-10 m/s^2` when moving upward ` v=0` t highest point a. `S=(v^2-u^2)/(2a) =((0.50)^2)/)2(-10))=125m ` maximum height reached = 125m b. `t= (v-u)/a=(0-50)/(-10)=5 sec c. S= 125/2= 62.5 m v=? u= 50 m/s a = -10 m/s^2 From v^2-u^2=2aS v=sqrt((u^2+2as)) =sqrt((50)^2+2(-10)(62.5)) =sqrt((2500-1250)) sqrt(1250)=35 m/s`. |
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