A projectile is fired at an angle of 55.0° above the horizontal with an initial speed of 35.0 m/s.
-What is the magnitude of the horizontal component of the projectile's displacement at the end of 2 s?
A)30 m
B)40 m
C)10 m
D)20 m
E)50 m

Question

Quizplus · Accepted Answer

The horizontal component of the initial velocity is given by $ v_{x} = v \cdot \cos(\theta) $. Here, $ v = 35.0 \, \text{m/s} $ and $ \theta = 55.0^\circ $. So, $ v_{x} = 35.0 \cdot \cos(55.0^\circ) \approx 20.1 \, \text{m/s} $. The horizontal displacement after 2 seconds is $ v_{x} \cdot t = 20.1 \, \text{m/s} \times 2 \, \text{s} \approx 40.2 \, \text{m} $, which is closest to 40 m.

A Projectile Is Fired at an Angle of 55