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16th March 2016, 02:46 PM
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Join Date: May 2012
Re: Annamalai University Engineering and Technology

The previous year question paper of Engineering Mechcanis subject of Department of Mechanical Engineering of University Annamalai under Engineering and Technology is as follows:

1.Boy ‘A ‘thrown a ball vertically up with a speed of 9 m/sec from the top of a shed of 2.5m
high . Boy ‘B” on the ground at the same instant throws a ball vertically up with a speed of 12
m/sec . Determine the time at which the two balls will be the same height above the ground.
What is the height?
Boy A: Consider the upward motion of the ball
u = 9 m/ Sec
a = - g = - 9.81 m/ sec2
t = time taken by the ball to reach the height “h”
SA = Height reached by the ball A
Equation of motion
+ ݐݑ = ݏ


ଶݐܽ
SA = 9 t + ½ (- 9.81) x t2
SA = 9 t - 4.905 t
…………(1)
Boy B : Consider the upward motion of the ball
u = 12 m/ sec
a = - g = - 9.81 m/ sec2
SB = Height reached by the ball B
SB = SA +2.5
t = time taken ( time remains same for both the cases)
Equation of motion
S = ut + ½ at2
SB = 12 t – ½ x 9.81 t2
= 12t – 4.905 t2 …………. (2)
Substitute (1) in (2)
SB = SA + 2.5 = 9 t - 4.905 t2
+ 2.5 = 12 t – 4.905 t

12t – 9 t = 2.5
t = 0.833 sec.
There fore
SA = 9 x 0.833 – 4.905 x 0.8332
= 4.093 m
SB = 4.093 + 2.5 = 6.593 m
Height of the two ball from ground = 6.593 m (Ans)




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