Question

g If you keep the launch angle fixed, but double the initial launch speed, what happens to the range?

Answer 1

Range will become 4 times of initial range

Step-by-step explanation:

Let the velocity of projection is u

And angle at which projectile is projected is
`\Theta`

And acceleration due to gravity is
`g\ m/sec^2`

So range of projectile is equal to
`R=(u^2sin2\Theta )/(g)`........eqn 1

Now in second case it is given that velocity of launching is doubled

So new velocity
`u_(new)=2u`

So new range will be equal to
`R_(new)=((2u)^2sin2\Theta )/(g)=(4u^2sin2\Theta )/(g)` .....eqn 2

Now dividing eqn 2 by eqn 1

`(R_(new))/(R)=(4u^2sin2\Theta )/(g)* (g)/(u^2sin2\Theta )`

`R_(new)=4R`

So if we double the initial launch speed then range will become 4 times