Question

can u please explain 42.
and 2nd part of 41.

Answer 1

42)

Initial velocity while thrown = v

So,

Initial velocity in vertical component while object is going upward

$= v_{o} \: \sin( \alpha )$

Final velocity of vertical component at top of trajectory = 0 m/s


Initial velocity of vertical component while coming downward =0 m/s

Final velocity of vertical component while coming downward
$= - v_{o} \: \sin( \alpha )$

So,

Total change in velocity of vertical component


$= ( - v_{o} \: \sin( \alpha ) ) - ( v_{o} \: \sin( \alpha ) ) \\ \\ = \bf \: - 2 v_{o} \: \sin( \alpha ) \: \: vertically \: \: upward \\ \\ = \bf \: 2 v_{o} \: \sin( \alpha ) \: \: vertically \: \: downward .$

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