SSC JE Mechanical Engineering 2017 Set 4 Paper

Question : 101 of 200

Marks: +1, -0

MECHANICAL

A body is thrown up at an angle of 45° with a velocity of 100 m /sec so as to describe a parabola. Its vertical velocity on point of return down will be

Zero
130 m/sec
50 m/sec
70.7 m/sec

Validate

Solution:

Let final velocity in vertical downward direction

b_y v_y.

vertical component of the velocity which goes particle up & down also.

∴ v_{oy} = v \sin 45°.

= 100 × sin 45° = 70.7 m/sec
option : ‘D’ = 70.7 m/sec

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