# Thread: Projectile Motion Question #2

1. ## Projectile Motion Question #2

I'm stuck on how to approach these 2 problems - any hints would be much appreciated!
1.
A body is projected form a point O on horizontal ground with velocity of 70m/s. It passes through a point P, which is 45m above the ground and 50m from O horizontally. Find the tangents of the two possible angles of elevation of the projected body. Find also the gradients at P of the two paths.
2.
The maximum range of a certain gun on a horizontal plane is R. Find the two possible angles at which the gun can be fired to give a range of 1/2R. If T1 and T2 are corresponding times of flight, find T1:T2.

2. Hello, ksinger!

I'll get you started on #1 . . .

$\text{1. A body is projected form a point }O\text{ on horizontal ground}$
$\text{with a velocity of 70m/s.\; It passes through a point }P\text{ which is}$
$\text{45m above the ground and 50m from }O\text{ horizontally.}$

$\text{(a) Find the tangents of the two possible angles of elevation}$
. . . . $\text{of the projected body.}$

We are expected to know these two parametric equations for projectile motion.

. . $\begin{array}{c}x \;=\;(70\cos\theta)t \qquad\quad \\ y \;=\;(70\sin\theta)t - 4.9t^2 \end{array} \quad\text{where }\theta\text{ is the angle of elevation}$

If $\,O$ is at the origin, then point $\,P$ is $(50,45).$

We have: . $x = 50\quad\Rightarrow\quad (70\cos\theta)t \:=\:50 \quad\Rightarrow\quad t \:=\:\dfrac{5}{7\cos\theta}$ .[1]
And: . $y = 45 \quad\Rightarrow\quad (70\sin\theta)t - 4.9t^2 \:=\:45$ .[2]

Substitute [1] into [2]:

. . $70\sin\theta\left(\dfrac{5}{7\cos\theta}\right) - 4.9\left(\dfrac{25}{49\cos^2\!\theta}\right) \;=\;45$

. . $50\tan\theta - 2.5\sec^2\!\theta \;=\;45 \quad\Rightarrow\quad 50\tan\theta - 2.5(\tan^2\!\theta + 1) \;=\;45$

. . $50\tan\theta - 2.5\tan^2\!\theta - 2.5 \;=\;45 \quad\Rightarrow\quad 2.5\tan^2\!\theta - 50\tan\theta + 47.5\;=\;0$

Multiply by 0.4: . $\tan^2\!\theta - 20\tan\theta + 19 \;=\;0$

Factor: . $(\tan\theta - 1)(\tan\theta - 19) \;=\;0$

. . $\text{Therefore: }\begin{Bmatrix}\tan\theta &=& 1 \\ \tan\theta &=& 19 \end{Bmatrix}$