Question

Two bullet are fired with horizontally velocities of 50m/ s and 100m/s from two guns at aheight of 19.6 .which bullet will strike first

Answer 1

oh........give him ...

Answer 2

Hope u like my process
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Time of flight is irrespective of velocity.

T =$\sqrt{ \frac{2h}{g} }$

So, both strikes at same time but what differs is range... i.e.

Range for this type of projectile is

$= > \bf \: r = v_{o} \sqrt{ \frac{2h}{g} } \\ where \: \: \: \: \: \: r = range \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: v_{o} = velocity \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: g = gravity \: \: (9.8 \: m {s}^{ - 1} ) \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: h = height$
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=> Height (h) = 19.6 m

So,

Range(R1) for bullet with velocity (50m/s) =

$= 50 \sqrt{ \frac{2 \times 19.6}{9.8} } = 50 \times 2 = \bf 100m$

Range(r2) for bullet with velocity(100m/s)=

$= 100 \sqrt{ \frac{2 \times 19.6}{9.8} } = 100 \times 2 = \bf200m$
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Hope this is ur required answer

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