Question

Suppose we replace the original launcher with one that fires the ball upward at twice the speed. We make no other changes. How far behind the cart will the ball land, compared to the distance in the original experiment?
(A) half as far
(B)twice as far
(C) four times as far
(D) the same distance
(E) by a factor not listed above

Answer 1

Answer:

(C) four times as far

Explanation:

As we know that the range of the launcher is given as

$R = \frac{v^2sin(2\theta)}{g}$

here we know that all the parameters will remain same but only change is the velocity is doubled

So we will have

$R' = \frac{(2v)^2 sin2\theta}{g}$

so we have

$R' = 4 R$

so here correct answer will be

(C) four times as far