Answer  Hint: In this given question of projectile, we have to use the relation for range and range depends only on velocity and angle of projection of projectile. Here the angle of projection is constant so range will only get affected due to velocity and it is directly proportional to velocity so range will increase if velocity will increase. Complete answer: It is given in the question that the initial velocity of the projectile to which it is fired is u.Angle of Projection is $\theta$.Range of the Projectile is R.These values are given in the question.Now we will apply the range formula for projectile. So this range of projectile can be expressed mathematically as,\[R=\dfrac{{{u}^{2}}Sin2\theta }{g}\] \[(Equation1)\].Now according to the question, the initial velocity of the projectile gets doubled.So let us assume this new velocity of projection is v.Therefore, \[v=2u\]\[(Equation2)\].Since , Angle of projection is same for this case so Range will only depend on velocity of projection of body. As Range is directly proportional to square of velocity of projection. So according to the question when velocity is doubled so its range of projectiles will also get four times. Now we should justify this result by doing some mathematical calculations.Let us assume new range of projectile will be R’ which can be mathematically expressed as following relation – \[{{R}^{'}}=\dfrac{{{v}^{2}}Sin2\theta }{g}\]Put the value of v from equation 2 in this relation then we get ,\[{{R}^{'}}=\dfrac{{{(2u)}^{2}}Sin2\theta }{g}\]on simplifying it we can obtain,\[\Rightarrow {{R}^{'}}=\dfrac{4{{u}^{2}}Sin2\theta }{g}\]\[(Equation3)\].Divide Equation 1 and Equation 3,we get\[\dfrac{{{R}^{'}}}{R}=\dfrac{\dfrac{4{{u}^{2}}Sin2\theta }{g}}{\dfrac{{{u}^{2}}Sin2\theta }{g}}\]on simplifying, we get \[\dfrac{{{R}^{'}}}{R}=4\]\[\therefore {{R}^{'}}=4R\]So the Range of the new projectile when the velocity of the projectile gets double is 4 times in comparison to the Range of initial projectile motion.Since we know that Range is the horizontal distance and this range of Projectile is maximum when angle of projection is \[{{45}^{0}}\]whatever the initial velocity of projection of the body.So , Correct Option is D. Note: Range is horizontal distance of projectile from point of projection to point of striking the ground. Its value can be calculated by multiplying the horizontal component of velocity and total time of flight. This horizontal component of velocity is constant throughout the motion and only the Vertical component of velocity will change throughout the motion.Why is it that if you double the velocity, the distance an object travels quadruples? from AskPhysics Text Solution Solution : Horizontal range, ` R=(u^(2) sin theta )g , i.e. R prop u^(@)` <br> When ` u' =2 `, then <br> ` R'/R=(2 u)^(@)/u^(@) =4 4 or R' =4 R` <br> It means if (u) is bounled, the horizontal range becomes four times the original horizontal range.
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What happens to range when velocity is doubled?
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