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Question

A ball is projected from the ground with velocity v such that its range is maximum
Column IColumn IIi. Velocity at half of the maximum height in vertical directiona.v2ii Velocity at the maximum heightb.v2iii. Change in its velocity when it returnsc.v2to the groundiv. Average velocity when it reaches d.v252the maximum height

A
i- a ii-b iii- d iv- c
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B
i- a ii-b iii- c iv- d
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C
i- b ii-a iii- d iv- c
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D
i- a ii-d iii- b iv- c
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Solution

The correct option is B i- a ii-b iii- c iv- d
i.a.,ii.b.,iiii.c.,iv.d.
i.a.
Range is maximum, when the angle of projection is 45
H=v22gsin245;H=v24g
Velocity, at half of the maximum height is v
v2=v2sin2452gH2;v2=v22v24
v2=v24v=v2
iib.
Velocity at the maximum height
v=v cos45v=v2
[because vertical component of velocity is zero at the highest point]
iiic.
Projection velocity
At projection point, vi=v cos45^i+v sin 45^j
At the point , when the body strikes the ground
vf=v cos45^iv sin45^j
Δv=vf+(vf)=2v sin 45(j)
|Δv|=2v sin 45=v2

ivd.Average velocity=Total displacementTotal time

Displacement=(R2)2+H2
vav=R24+H2vsinθg=R2+4H22vsinθg
vav=(v2g)2+4(v24g)22vg
=v2g1+14v2g=v252v2=v252

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