A projectile is launched from the edge of a cliff with a velocity of 45.0 m/s at an angle of 30.0 degrees above the horizontal. Relative to its starting point, what is the location of the projectile 3.00 seconds later?
A) X = 85.0 m, Y = 37.0 m
B) X = 117.0 m, Y = 23.4 m
C) X = 37.0 m, Y = 117.0 m
D) X = 23.4 m, Y = 37.0 m
E) X = 85.0 m, Y = 23.4 m

Question

Quizplus · Accepted Answer

First, we need to separate the horizontal and vertical components of the initial velocity. 
Vy = V * sin(30) = 45 / 2 = 22.5 m/s (upward)
Vx = V * cos(30) = 45 * √3 / 2 = 38.9 m/s (horizontal)
Next, we can use the kinematic equations to find the location of the projectile after 3 seconds. 
For vertical motion: 
Y = Voy*t + 1/2 * a * t^2
Y = 22.5 * 3 - 1/2 * 9.81 * 3^2
Y = 23.4 m (upward displacement)
For horizontal motion: 
X = Vx * t
X = 38.9 * 3
X = 117.0 m (horizontal displacement)
Therefore, the location of the projectile relative to its starting point after 3 seconds is X = 117.0 m, Y = 23.4 m. Choice B is the best.

A Projectile Is Launched from the Edge of a Cliff