A stone is thrown with an initial speed of 11.5 m/s at an angle of 50.0 above the horizontal from the top of a 30.0-m-tall building. Assume air resistance is negligible, and g = 9.8 m/s2.
-What is the magnitude of the horizontal displacement of the rock?
A)3.53 m
B)13.3 m
C)222 m
D)26.1 m

Question

Quizplus · Accepted Answer

The horizontal displacement of a projectile can be found using the formula $d = v \cos(\theta) \times t$, where $v$ is the initial velocity, $\theta$ is the launch angle, and $t$ is the time of flight. To find $t$, we use the vertical motion equation $y = v \sin(\theta) t - \frac{1}{2} g t^2$, where $y$ is the vertical displacement. Since the stone is thrown from a height, we set $y = -30.0 \, \text{m}$ (taking downward as negative), $v = 11.5 \, \text{m/s}$, $\theta = 50^\circ$, and $g = 9.8 \, \text{m/s}^2$. Solving for $t$ gives the time of flight, which can then be used to calculate the horizontal displacement. The correct calculation leads to a displacement of approximately 26.1 meters, making option D the correct answer.

A Stone Is Thrown with an Initial Speed of 11