Question

Find the velocity of the projection of a bullet fired at an angle of 15 degree if it hits the ground 5 kilometre away.

Answer 1

Range R=u  

2

Sin2θ/g=6000meter

so u  

2

=120000m/s so u=346.4m/s

Note, we used Sin2θ=Sin30  

0

=0.5

Now we know that the maximum value of range is given as R  

max

​  

=u  

2

/g=120000/10=12000meter=12Km

So its possible to hit the target at 10Km  distance because its less than maximum range.