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(problem reference 3 - 2) a projectile fired from a gun has initial horizontal and vertical components of velocity equal to 30 m/s and 40 m/s, respectively.
at what angle is the projectile fired (measured with respect to the horizontal)?
a. 37°
b. 53°
c. 45°
d. 59°

Explanation:

Step1: Recall tangent - angle relation

The tangent of the angle $\theta$ of projection with respect to the horizontal is given by $\tan\theta=\frac{v_y}{v_x}$, where $v_y$ is the initial vertical - component of velocity and $v_x$ is the initial horizontal - component of velocity.
Given $v_x = 30$ m/s and $v_y=40$ m/s, so $\tan\theta=\frac{40}{30}=\frac{4}{3}$.

Step2: Calculate the angle

We know that if $\tan\theta=\frac{4}{3}$, then $\theta=\arctan(\frac{4}{3})$.
Using a calculator, $\theta\approx53^{\circ}$ (since $\arctan(\frac{4}{3})\approx53.13^{\circ}\approx53^{\circ}$).

Answer:

B. $53^{\circ}$