Question

a canon ball was fired at an angle of 30 degrees with the height of 50m above the ground. If the initial horizontal velocity is 100 m/s determine the ff.

a.how long it will stay in the air

b.the horizontal distance that it will travel

c.finds its final velocity

Answer 1

Let's first find the initial velocity of the canon ball:

$$v_x=v_0cos\theta,$$ $$v_0=\dfrac{v_x}{cos\theta}=\dfrac{100\ \dfrac{m}{s}}{cos30^{\circ}}=115.5\ \dfrac{m}{s}.$$

(a) We can find the time flight of the canon ball from the kinematic equation:

$$y=y_0+v_{0y}t-\dfrac{1}{2}gt^2,$$ $$0=50+115.5\cdot sin30^{\circ}t-\dfrac{1}{2}\cdot9.8t^2,$$ $$4.9t^2-57.75t-50=0.$$

This quadratic equation has two roots: $t_1=12.6\ s$ and $t_2=-0.81\ s$. Since time can't be negative the correct answer is $t=12.6\ s$.

(b) We can find the horizontal distance that it will travel from the kinematic equation:

$$x=v_0tcos\theta=115.5\ \dfrac{m}{s}\cdot12.6\ s\cdot cos30^{\circ}=1260\ m.$$

(c) The initial horizontal velocity remains unchanged during the flight. Let's find the vertical velocity of the canon ball:

$$v_y=v_0-gt=0-9.8\ \dfrac{m}{s^2}\cdot12.6\ s=-123.48\ \dfrac{m}{s}.$$

The sign minus means that the vertical velocity directed downward.

Finally, we can find the final velocity of the canon ball from the Pythagorean theorem:

$$v=\sqrt{v_x^2+v_y^2}=\sqrt{(100\ \dfrac{m}{s})^2+(-123.48\ \dfrac{m}{s})^2}=159\ \dfrac{m}{s}.$$