Question

A truck is moving horizontally with velocity V. A ball is thrown with velocity U at an angle o above the
horizontal from the truck relative to truck. The horizontal range of the shell relative to the ground is
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Answer 1

Answer:

Answer: The shell will have two component of velocity v being horizontal and u being vertical as seen by ground observer

so the range will be =v*t

=v*time

=v*2u/g

Answer 2

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Here an important point to keep in mind is that shell makes angle B wrt vertical , hence wrt horizontal , the horizontal component for shell will be

ucos(90- B)

= usinB

Velocity of car is given as v

Hence total horizontal velocity component = v+usinB

Let the time taken by the projectile motion be t.

When the shell reaches ground, the vertical displacement is 0.

Also from 2nd equatin of motion , we know that,

y = utcosB – ½ gt2

we are considering vertical component of velocity of the shell which comes out to be

usin(90-B)

we have kept in mind here that the car does not have any vertical component of velocity.

Putting y = 0 in 2nd equation of motion, we get:

0 = ucosB – ½ gt

Or t = 2ucosB/g

Range now is nothing but the horizontal distance travelled which will be the product of horizontal component of velocity and time.

Hence range = (v+usinB)t

Substituting the value of t from above we get:

(v + usinB)(2ucosB)/g

=(2vucosB +u²sin2B)/g which is the required answer.

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