Answer to Question #274512 in Physics for roy

Question

What are the horizontal range and maximum height of a bullet fired with a speed of 20 m/s at 30° above the horizontal?

Answer 1

(a) Let's first find the time that the bullet takes to reach the maximum height:

$$v=v_0sin\theta-gt,$$ $$0=v_0sin\theta-gt,$$ $$t=\dfrac{v_0sin\theta}{g}.$$

Then, we can find the flight time of the bullet:

$$t_{flight}=2t=\dfrac{2v_0sin\theta}{g},$$ $$t_{flight}=\dfrac{2\times20\ \dfrac{m}{s}\times sin30^{\circ}}{9.8\ \dfrac{m}{s^2}}=2.04\ s.$$

Finally, we can find the horizontal range of the bullet from the kinematic equation:

$$x=v_0 t_{flight}cos\theta,$$ $$x=20\ \dfrac{m}{s}\times2.04\ s\times cos30^{\circ}=35.3\ m.$$

(b) We can find the maximum height reached by the bullet from the kinematic equation:

$$y_{max}=v_0tsin\theta-\dfrac{1}{2}gt^2.$$

Substituting $t$ into the previous equation, we get:

$$y_{max}=v_0sin\theta(\dfrac{v_0sin\theta}{g})-\dfrac{1}{2}g(\dfrac{v_0sin\theta}{g})^2,$$ $$y_{max}=\dfrac{v_0^2sin^2\theta}{2g},$$ $$y_{max}=\dfrac{(20\ \dfrac{m}{s})^2sin^230^{\circ}}{2\times9.8\ \dfrac{m}{s^2}}=5.1\ m.$$