Question

A projectile is fired at an angle of 30 degrees above the horizontal from the top of a cliff 600 ft high. The initial speed of the projectile is 2000 ft/s

a. Find the range of projectile
b. Find the time of flight

Answer 1

Let's first write the projections of initial velocity of the projectile on axis x- and y:

$$v_{x} = v_0cos\theta,$$ $$v_y = v_0sin\theta,$$

here, $v_0 = 2000 \ \dfrac{ft}{s}$ is the initial velocity of the projectile, $\theta = 30^{\circ}$ is the launch angle.

Let's write the equations of the projectile motion along axis $x$- and $y$ (also, we choose the upwards as the positive direction):

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

here, $x$ is the range of the projectile, $y = -600 \ ft$ is the vertical displacement of the projectile (or the height), $t$ is the time of flight, $g = -32 \ \dfrac{ft}{s^2}$ is the acceleration due to gravity.

We can find the time of flight from the second equation:

$$-600 = 2000 \cdot sin30^{\circ} \cdot t - \dfrac{1}{2} \cdot 32 \cdot t^2,$$ $$16t^2 - 1000t -600 = 0.$$

This quadratic equation has two roots:

$$t_1 = \dfrac{-(-1000)+ \sqrt{(-1000)^2-4 \cdot 16 \cdot (-600)}}{2 \cdot 16} = 63 \ s,$$ $$t_2 = \dfrac{-(-1000)- \sqrt{(-1000)^2-4 \cdot 16 \cdot (-600)}}{2 \cdot 16} = -0.6 \ s.$$

Since time can't be negative, the correct answer is $t = 63 \ s$.

Finally, we can find the range of projectile from the first equation:

$$x = 2000 \ \dfrac{ft}{s} \cdot cos30^{\circ} \cdot 63 \ s = 109116 \ ft.$$

Answer:

a) $x = 109116 \ ft.$

b) $t = 63 \ s.$