A projectile is thrown from the top of a tall building with a velocity of 15.0 m/s at an angle of 30.0 degrees above the horizontal. Relative to its starting point, what is the location of the projectile 2.00 seconds later?
A) X = 4.60 m, Y = 9.60 m
B) X = −26.0 m, Y = −26.0 m
C) X = −36.0 m, Y = −4.60 m
D) X = 26.0 m, Y = −4.60 m
E) X = 46.0 m, Y = 4.60 m

Question

Quizplus · Accepted Answer

To find the location of the projectile after 2.00 seconds, we can break the motion into horizontal (x) and vertical (y) components. The initial velocity can be decomposed into $v_{x0} = v_0 \cos(\theta)$ and $v_{y0} = v_0 \sin(\theta)$, where $v_0 = 15.0 \, \text{m/s}$ and $\theta = 30.0$ degrees.For the horizontal motion, since there's no acceleration (assuming no air resistance), $x = v_{x0}t$. With $v_{x0} = 15.0 \cos(30.0^\circ) \approx 12.99 \, \text{m/s}$, and $t = 2.00 \, \text{s}$, we get $x \approx 25.98 \, \text{m}$, which rounds to 26.0 m.For the vertical motion, we use $y = v_{y0}t - \frac{1}{2}gt^2$, where $g = 9.8 \, \text{m/s}^2$. With $v_{y0} = 15.0 \sin(30.0^\circ) = 7.5 \, \text{m/s}$, and $t = 2.00 \, \text{s}$, we get $y = 7.5(2.00) - 0.5(9.8)(2.00)^2 \approx -4.60 \, \text{m}$.Thus, the projectile's location relative to its starting point is $X = 26.0 \, \text{m}$, $Y = -4.60 \, \text{m}$, which matches choice D.

A Projectile Is Thrown from the Top of a Tall