Question

1. A bullet is fired at the center of a target that is 103 meters away. If the bullet has an initial velocity of 810 m/s, how far below the center of the target will the bullet be when it hits the target? Take gravitational acceleration to be 10 m/s².
(Hint: How long will it take for the bullet to reach the target?)

2. A ball rolls off a table with a horizontal velocity of 8 m/s.  If the table is 0.76 meters tall, how far away from the table will it land?  Take acceleration due to gravity to be at 10 m/s².
(Hint:  How long does the ball take to fall to the ground?)

Answer 1

Answers:

1), The bullet will be 0.085 m below the centre of target

2)The ball will land 0.48 m away from table

Explanation:

1)Distance =103 m

Velocity of the bullet = 810 m/s

we have ,

Distance = velocity * time

or , time = distance/speed

Here, Time taken for bullet to reach the target ,

t = 103/810 = 0.13 s

We have kinematics equation for distance covered by bullet,

s= ut +(1/2)at²

To find how far below the centre of target the bullet be, we take the vertical components:

s_y= u_yt +(1/2)a_yt²

u_y= vertical component of velocity of bullet = 0 because the bullet is travelling in horrizontal direction towards the target.

a_y=g= acceleration due to gravity = 10 m/s²

t =0.13 s ( we found above)

Substituting in s_y,

s_y= 0+(1/2)*10*0.13²

= 0.085 m

So, the bullet will be 0.085 m below the centre of target .

2)Given,

height of table = 0.76 m

velocity of ball =8 m/s

Time taken for ball to fall to ground= Distance/velocity

t = 0.76/8 =0.095 s

We have ,

s = ut +(1/2) at²

Considering the vertical components,

s_y= u_yt +(1/2)a_yt²

u_y= 0

a_y= g = 10 m/s

t= 0.095 s

Substituting,

s_y= 0+(1/2)*10*0.095 =0.48 m

So ,the ball will land 0.48 m away from table.