Answer to Question #179553 in Physics for Nicole

Question

Two projectiles, P1 and P2, are fired at different angles from a common starting line and ultimately land the same distance away on level ground. If P1 is launched at 25.0° from horizontal, what can be concluded about P2's angle of launch from horizontal, time in air, and speed at which it lands compared to P1?

A) launch angle= 45.0°, time in air= less than P1, landing speed= same as P1

B) launch angle= 45.0°, time in air= same as P1, landing speed= zero

C) launch angle= 65.0°, time in air= greater than P1, landing speed= same as P1

D) launch angle= 65.0°, time in air= less than P1, landing speed= greater than P1

Answer 1

The horizontal distance covered by a projectile is given as follows:

$$d = \dfrac{v^2\sin^2\theta}{2g}$$

where $v$ is the launch speed (which is equal to landing speed), $\theta$ is the launch angle, $g$ is the gravitational acceleration. Since the distance is the same for both P1 and P2, obtain:

$$\dfrac{v_1^2\sin^2\theta_1}{2g} = \dfrac{v_2^2\sin^2\theta_2}{2g}\\ v_1^2\sin^2\theta_1 = v_2^2\sin^2\theta_2$$

Expressing the landing speed of P2, obtain:

$$v_2 = v_1\dfrac{\sin\theta_1}{\sin\theta_2}$$

The launch angle of P1 is $\theta_1 = 25\degree$. Suppose $\theta_2 = 45\degree\space or\space \theta_2 = 65\degree$. In any case ratio $\dfrac{\sin\theta_1}{\sin\theta_2}>1$ and the speed of P1 is greater than the speed of P2.

Answer. D.