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
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:
- 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)
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)