InterviewSolution
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InterviewSolution
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A ball is thrown up at a speed of 4.0 m/s. Find the maximum height reached by the ball. Take `g=10 m//s^2`. |
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Answer» Let us take vertically upward direction as the positive Y-axis. We have `u=4.0 m/s and a=-10m/s^2`. At the highest point the velocity becomes zero.Using the formula. `v^2=u^2+2ay`, `0=(4.0 m/s)^2+2(-10 m/s^2)y` ltbrlt or, `y=(16 m^2/s^2)/(30 m/s^2)=0.80 m`. |
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