 or  # A ball is thrown upward with an ini tial velocity of 100 "ms"^(-1). After how much time will it return ? Draw velocity - time graph for the ball and find from the graph (i) the maximum height attained by the ball and (ii) height of the ball after 15 s. Take g=10 ms^(-2) . Apne doubts clear karein ab Whatsapp par bhi. Try it now.
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Solution :
<img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/AKS_CAO_PHY_V01_P1_C03_SLV_042_S01.png" width="80%"> <br> Here , `u=100 ms^(-1), g=-10 ms^(-2)` <br> At highest point , v=0 <br> As v=u+gt `therefore` 0=100-10 x t <br> `therefore` Time taken to reach highest point , `t=100/10`=10s <br> The ball will return to the ground at t = 20 s <br> Corresponding velocity- time graph of the ball is shown in <br> (i)Maximum height attained the ball=Area of `triangleAOB` <br> `=1/2xx10xx100` =500 m <br> (ii)Height attained after 15s=Area of `triangleAOB`+Area of `triangleBCD` <br> `=500+1/2(15-10)xx(-50)=500-125`=375 m 