Question

a projectile is launched from level ground at an angle θ < 45° above the horizontal. the projectile lands some distance away with speed v. if both the speed and the angle with which the projectile was launched were doubled, what would the new speed be when it lands?
a v
b 2v
c 4v sin θ
d 4v

Explanation:

Step1: Recall projectile - motion property

The speed of a projectile when it lands on level - ground is equal to its initial speed due to conservation of mechanical energy (since the initial and final heights are the same, so the change in gravitational potential energy is zero). Let the initial speed be $u$. The speed when it lands is $v = u$.

Step2: Consider the new initial conditions

If the initial speed $u$ is doubled to $u'=2u$ and the launch angle $\theta$ is doubled to $\theta' = 2\theta$, the speed when it lands is still equal to the initial speed because of energy conservation. Since $u' = 2u$ and originally $v = u$, the new speed $v'$ when it lands is $v'=2u$. But since $v = u$, the new speed $v'=2v$.

Answer:

B. $2v$