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A stone is thrown vertically up with a speed of 30m/s from a very tall cliff from its edge . Find the average velocity over 10 seconds from the instant it was thrown up. g=10 m/s^2

A stone is thrown vertically up with a speed of 30m/s from  a very tall cliff from its edge . Find the average velocity over 10 seconds from the instant it was thrown up. g=10 m/s^2

Grade:11

2 Answers

sunny patel
13 Points
6 years ago
Stone will first reach a maximum height in time ‘t’. then it will descend to the point of throw in another time ‘t’ with velocity 30m/s downwards. ‘t’ can be calculated from v = u + at. So ‘t’ = 3sec. So in total 6sec the stone is at the point of start i.e no displacement at all. Now we have to calculate distance covered byy stone in next 4 seconds from s = ut + 1/2(at^2). So distance traveled in next 4 seconds comes to be 200m. In terms of displacement the stone’s displacement is 200m in 10 seconds. therefore average velocity is equal to 200m/10sec = 20m/s.
Vikas TU
14149 Points
6 years ago
Dear Student,

Average Velocity=Total Displacement/Total time

H=Displacement
From the equation of motion,
S=ut+at2/2
=>H=30x10-g102/2
=>H=-200m

the –ve sign here signifies that the ball has descended 200 m below from its point of projection. (normal sign convention)

Hence magnitude of displacement=200m
Time=10s
Average Velocity=200/10=20m/s
Cheers!!
Regards,
Vikas (B. Tech. 4th year
Thapar University)

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