#1
16th March 2016, 02:43 PM
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Annamalai University Engineering and Technology
Can you provide me previous year question paper of Engineering Mechcanis subject of Department of Mechanical Engineering of University Annamalai under Engineering and Technology?
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#2
16th March 2016, 02:46 PM
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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 ଵ ଶ 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) |