UPSEE Solved Model Paper 7

Given:
Initial horizontal velocity of bullet, v = 1000 m/s
Distance from the target (a bird), d = 100 m [horizontal distance]
Let t be the time it takes to hit the target. Then,
t =

d

v

=

100m

1000m∕s

= 0.1 s
[Note: There is no force acting on the bullet in horizontal direction. Thus horizontal velocity remains constant.]
During this time, the bullet will not only cover a horizontal distance of 100 m but also will fall vertically downward (because of acceleration due to gravity)
through a height say h.
Now, initial vertical velocity of bullet is zero. Using second equation of motion:
S = ut+

1

2

at2 , we have
h = (0 × t) +

1

2

gt2
⇒ h =

1

2

×10m∕s2×(0.1s)2
⇒ h = 0.05 m