A ball is thrown upward with an initial velocity of 100 ms^(-1). After how much time will it return ? Draw velocity -time graph for the ball and find the maximum height attained by the ball.[Assume, g = - 10 ms ^ { - 2 }]

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

1.	A ball is thrown upward with an initial velocity of 100 ms^(-1). After how much time will it return ? Draw velocity -time graph for the ball and find the maximum height attained by the ball.[Assume, g = - 10 ms ^ { - 2 }]
Answer» Solution :Given : ` u = 100 ms^(-1) , g = - 10 ms^(-2)` At highest point , v = 0 As v = u + g t `:. 0 = 100 - 10 xx t` `:.`Time TAKEN to reach highest point, `t = (100)/(10) = 10 s` The ball will return to the ground at t = 20 s Velocities of the ball at different instants at time will be as follows : At `t = 0, v = 100 - 10 xx 0 = 100 ms^(-1)` At ` t = 5 s, v = 100 - 10 xx 5 = 50 ms^(-1)` At ` t = 10 s, v = 100 - 10 xx 10 = 0` At ` t = 15 s, v = 100 - 10 xx 15 = - 50 ms ^(-1)` At `t = 20 s, v = 100 - 10 xx 20 = - 100ms^(-1)` The VELOCITY - time graph will be as shown in the figure . Height attained after 15 s = AREA of `Delta AOB` + Areaof`Delta BCD` `= 500 +(1)/(2) (15 - 10) xx (-50)` = 500 - 125 = 375 m