Answer to Question #137063 in Physics for JJ Thompson

Question

An artillery is pointed upward at an angle of 30 degrees with respect to the horizontal with
a muzzle velocity of 1500 ft/s. Ignoring the effect of air resistance, to what height will the
projectile rise and what will be its range?

Answer 1

We can find the maximum height that the projectile can reach from the kinematic equation:

$$v_{y,f}^2=v_{y,i}^2 - 2ay_{max},$$

here, $v_{y,f} = 0 \ \dfrac{ft}{s}$ is the final velocity of the projectile at a maximum height, $v_{y,i} = v_isin \theta$ is the vertical component of the velocity of the projectile, $v_i = 1500 \ \dfrac{ft}{s}$ is the initial velocity of the projectile, $\theta = 30^{\circ}$ is the launch angle, $a = g =32 \ \dfrac{ft}{s^2}$ is the acceleration due to gravity, $y_{max}$ is the maximum height that the projectile can reach.

Then, from this equation we can find $y_{max}$:

$$y_{max} = \dfrac{(v_isin\theta)^2}{2g} = \dfrac{(1500 \ \dfrac{ft}{s} \cdot sin30^{\circ})^2}{2 \cdot 32 \ \dfrac{ft}{s^2}} = 8789 \ ft.$$

We can find the range of the projectile from the formula:

$$R = \dfrac{v_i^2sin2\theta}{g} = \dfrac{(1500 \ \dfrac{ft}{s})^2 \cdot sin2\cdot 30^{\circ}}{32 \ \dfrac{ft}{s^2}} = 60892 \ ft.$$

Answer:

$y_{max} = 8789 \ ft, R = 60892 \ ft.$