Question

a bullet is fired at an angle of 45 degree with the horizontal with a velocity of 49 ms^-1.find the time of fight and horizontal range​

Answer 1

This is projectile motion, in which only the vertical component of initial velocity keeps changing, but the horizontal doesn't change at all.

Let the initial velocity be u.

Horizontal range for 45

O

$$projection=\cfrac{u^2}g=360m$$

∴u=

10

360

=6m/s

Time taken for entire flight is T=

g

2usinθ

=

10

2

×6

sec.

18km/hr=

3600

18000

=5m/s

Distance travelled by lorry =5×T=

2

6

=4.243 metres

This distance will get added up to the old range.

New range =364.243 m.

Answer 2

Answer:

bhai yoir answer is

me aur kuch help kar sakta hu kya