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प्रश्न
A ball is thrown up vertically and returns back to thrower in 6 s. Assuming there is no air friction, plot a graph between velocity and time. From the graph calculate
- deceleration
- acceleration
- total distance covered by ball
- average velocity.
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उत्तर
A ball is thrown up vertically and returns to thrower in 6s. It means the ball takes 3s to reach the highest point and 3s to reach the earth from the highest point.
For the upward motion of a ball
u = ?
v = 0
a = −g = −10 ms−2
t = 3s
v = u + at
0 = u − 10(3)
u = 30 m/s
⇒ In the graph, OA = CD = 30 m/s
(i) When the ball moves upwards, then it decelerates.
From graph, deceleration = − slope of graph AB
= `-"OA"/"OB"=-30/3` = −10 ms−2
(ii) When ball falls downwards then acceleration = Slope of v.t. graph BC
= `"CD"/"BD"=30/3` = 10 ms−2
(iii) Total distance covered = area under v.t. graph
= ar(ΔOAB) + ar(ΔBCD)
= `1/2xx"OB"xx"OA" + 1/2xx"BD"xx"CD"`
= `1/2xx30xx3+1/2xx30xx3`
= 45 + 45
= 90 m
(iv) Ball returns back to thrower in 6s.
⇒ Displacement after 6s = 0
Average velocity = `"Time distance covered"/"Total time taken"=0/6` = 0
Average speed = `"Total distance covered"/"Total time taken"=90/6` = 15 ms−1
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