A boy throws a ball at an angle of 60 degrees above the horizontal with a speed of 12.00 m/s. After 2 s, what is the distance of the ball from the boy?
A)1.18 m
B)10.82 m
C)12.06 m
D)24.0 m

Question

Quizplus · Accepted Answer

The distance of the ball from the boy can be found using the formula for the range of a projectile: $R = \frac{v^2 \sin 2\theta}{g}$, where $v$ is the initial velocity, $\theta$ is the launch angle, and $g$ is the acceleration due to gravity (approximately $9.8 m/s^2$). However, this specific question asks for the distance after 2 seconds, not the total range. To find this, we need to calculate the horizontal and vertical components of the initial velocity and then use them to find the position after 2 seconds. The horizontal component of the velocity ($v_x$) is $v \cos \theta = 12 \cos 60^\circ = 6 m/s$, and the vertical component ($v_y$) is $v \sin \theta = 12 \sin 60^\circ = 10.39 m/s$. After 2 seconds, the horizontal distance ($x$) traveled is $v_x \times t = 6 \times 2 = 12 m$, and the vertical distance ($y$) can be found using $y = v_y t - \frac{1}{2} g t^2 = 10.39 \times 2 - \frac{1}{2} \times 9.8 \times 2^2 = 20.78 - 19.6 = 1.18 m$. However, the question asks for the straight-line distance from the boy, which can be found using the Pythagorean theorem: $\sqrt{x^2 + y^2} = \sqrt{12^2 + 1.18^2} \approx 12.06 m$.

A Boy Throws a Ball at an Angle of 60