A ball is thrown upward with an initial velocity of 100ms^(-1). After how much time will it return? Draw velocity -time graph for the ball and find the graph (i) the maximum height attained by the ball and (ii) height of the ball after 15s. Take g=10ms^(-2).

A ball is thrown upward with an initial velocity of 100ms^(-1). After

1.	A ball is thrown upward with an initial velocity of 100ms^(-1). After how much time will it return? Draw velocity -time graph for the ball and find the graph (i) the maximum height attained by the ball and (ii) height of the ball after 15s. Take g=10ms^(-2).
Answer» Solution : Here, `u=100ms^(-1)""g=-10ms^(-2)` At HIGHEST point, v=0 As `v=u+gt therefore 0=100 -10 xx t` Time taken to reach highest point, `t=(100)/(10)=10S` The ball will return to the GROUND at t=20s Corresponding velocity -time graph of the ball is shown in (i) Maximum height attained the ball =Area of `triangleAOB=1/2 xx 10 xx 10=500m` (ii) Height attained after 15s= Area of `triangleAOB+"Area of "triangleBCD` `=500+1/2(15-10)xx (-50)=500-125=375m`