Answer to Question #285861 in Physics for Junard

Question

A projectile is fired into the air from the top of 200-m cliff above a valley. Its initial velocity is 60m/s at 60° above the horizontal. Where does the projectile land? Show the given,unknown,and solution.

Answer 1

Given: $y_0=200\ m$ is the initial height, $v_0=60\ \dfrac{m}{s}$ is the initial velocity, $\theta=60^{\circ}$ is the launch angle.

Unknown: $x$, the range of the projectile.

Solution:

Let's first find the total time in air from the kinematic equation:

$$y=y_0+v_{0y}t+\dfrac{1}{2}gt^2,$$ $$0=200+60t\times sin60^{\circ}+\dfrac{1}{2}\times(-9.8)t^2,$$ $$4.9t^2-52t-200=0.$$

This quadratic equation has two roots: $t_1=13.6\ s$, $t_2=-2.99\ s$. Since time can't be negative the correct answer is $t=13.61\ s$.

Finally, we can find the range of the projectile as follows:

$$x=v_{0x}t=v_0tcos\theta,$$ $$x=60\ \dfrac{m}{s}\times 13.6\ s\times cos60^{\circ}=408\ m.$$