A projectile is shot from the edge of a vertical cliff 60.0 m above the ocean. It has a speed of 100 m/s and is fired at an angle of 35.0° below the horizontal. How far from the foot of the vertical cliff does the projectile hit the water?
A) 79.1 m
B) 312 m
C) 684 m
D) 1084 m
E) 540 m

Question

Quizplus · Accepted Answer

The correct answer is A because the projectile's horizontal displacement is given by:$$x = v_0 \cos(	heta) t$$where $$v_0$$ is the initial velocity, $$	heta$$ is the angle below the horizontal, and t is the time of flight. The time of flight is given by:$$t = \frac{2v_0 \sin(	heta)}{g}$$where g is the acceleration due to gravity. Substituting these expressions into the equation for x, we get:$$x = v_0 \cos(	heta) \frac{2v_0 \sin(	heta)}{g}$$$$x = \frac{v_0^2 \sin(2	heta)}{g}$$Plugging in the given values, we get:$$x = \frac{(100 \ m/s)^2 \sin(2(35.0°))}{9.81 \ m/s^2}$$$$x = 79.1 \ m$$

A Projectile Is Shot from the Edge of a Vertical