Question

A cannon ball is fired with an initial velocity of 25m/s at 53º from the horizontal. The cannon is
situated on a cliff 60 m high. Find the (a) maximum height the cannon ball reached; (b) the
maximum horizontal distance it reached; (c) total time the cannon ball is in the air.

Answer 1

a) Let's write the equations of motion of the cannon ball in horizontal and vertical directions:

$$x=v_0tcos\theta, (1)$$ $$y=y_0+v_0tsin\theta-\dfrac{1}{2}gt^2, (2)$$

here, $x$ is the horizontal displacement of the cannon ball (or the range of the cannon ball), $v_0=25.0\ \dfrac{m}{s}$ is the initial velocity of the cannon ball, $t$ is the total flight time of the cannon ball, $\theta=53^{\circ}$ is the launch angle, $y$ is the vertical displacement of the cannon ball (or the height), $y_0=60.0\ m$ is the initial height from which the cannon ball was fired and $g=9.8\ \dfrac{m}{s^2}$ is the acceleration due to gravity.

Let's first find the time that the cannon ball takes to reach the maximum height from the kinematic equation:

$$v_y=v_0sin\theta-gt_{rise},$$ $$0=v_0sin\theta-gt_{rise},$$ $$t_{rise}=\dfrac{v_0sin\theta}{g}.$$

Then, we can substitute $t_{rise}$ into the second equation and find the maximum height:

$$y_{max}=y_0+v_0sin\theta\cdot\dfrac{v_0sin\theta}{g}-\dfrac{1}{2}g(\dfrac{v_0sin\theta}{g})^2,$$ $$y_{max}=y_0+\dfrac{v_0^2sin^2\theta}{2g},$$ $$y_{max}=60\ m+\dfrac{(25.0\ \dfrac{m}{s})^2\cdot sin^253^{\circ}}{2\cdot 9.8\ \dfrac{m}{s^2}}=80.3\ m.$$

(b)-(c) Let's first find total flight time of the cannon ball:

$$y=y_0+v_0tsin\theta-\dfrac{1}{2}gt^2,$$ $$0=60+25tsin53^{\circ}-\dfrac{1}{2}\cdot 9.8t^2,$$ $$4.9t2-20t-60=0.$$

This quadratic equation has two roots:

$$t_1=\dfrac{-b+\sqrt{b^2-4ac}}{2a}=\dfrac{20+\sqrt{(-20)^2+1176}}{2\cdot 4.9}=6.09\ s,$$ $$t_2=\dfrac{-b-\sqrt{b^2-4ac}}{2a}=\dfrac{20-\sqrt{(-20)^2+1176}}{2\cdot 4.9}=-2.01\ s.$$

Since time can't be negative, the correct answer is $t=6.09\ s.$

Finall, we can substitute $t$ into the first equation and find the maximum horizontal distance reached by the cannon ball:

$$x=v_0tcos\theta=25.0\ \dfrac{m}{s}\cdot 6.09\ s\cdot cos53^{\circ}=91.6\ m.$$

Answer:

(a) $y_{max}=80.3\ m.$

(b) $x=91.6\ m.$

(c) $t=6.09\ s.$