Question
Wed June 13, 2012 By: Ratik Mahajan
A ball is thrown upwards with an initial velocity of 100 m/s. After how much time will it return? Draw a velocity-time graph for the ball and find from the graph-
(i)maximum height attained by the ball.

A ball is thrown upwards with an initial velocity of 100 m/s. After how much time will it return? Draw a velocity-time graph for the ball and find from the graph- (i)maximum height attained by the ball. (ii)height of the ball after 15 seconds. Take g= 10

Expert Reply
Thu June 14, 2012

For answering the question we have to assume the sign convention of our Cartesian frame as up direction as positive.

The initial velocity of ball is given as:   u= 100m/s

The acceleration due to gravity is:        g= -10m/s2 (negative sign is because of it   being acting in downward direction)

Let Displacement after time ‘t’ be ‘s’.

Using 2nd equation of motion:

s=0 for final displacement being 0 i.e. the ball has returned back.

So that t=0 for throwing time or t=20s, the time taken by ball to return.

Let the velocity of that ball after time ‘t’ be ‘v’.

Using 1st equation of motion:

Plotting we get:

  1. Maximum height is attained when the velocity of ball becomes 0m/s at the highest point i.e. the area under the graph between time 0 to 10 (as displacement can be shown as area under the velocity-time curve because displacement= velocity*time)

        So that:

       

        s=500m

    2. After 15s the height will be

      

            So that s= 500-125 = 375m

       (Velocity after 10 seconds is taken as negative because the ball is now coming downwards)

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