1. ## [SOLVED] Projectiles....

A stone is projected from a point O on a cliff with a speed of 20 m/s at an angle of elevation of 30(degrees) . T seconds later the angle of depression of the stone from O is 45(degrees). Find the value of T .

i encountered this problem while solving M2 past exam quetions and i couldnt solve it... ... to make things worse no one in my class knows how to solve this problem . Pls hlp me cuz xam is in a few days. i will be very grateful...

2. I am including a diagram. I hope I am interpreting correctly.

We can use the horizontal and vertical component formulas.

horizontal: $\displaystyle x=(20cos(30))t$

vertical: $\displaystyle y=(20sin(30))t-\frac{1}{2}(9.8)t^{2}$

When the rock is level with the point it was thrown, then it's the point where y equals 0, we can find x at that time.

$\displaystyle 0=20sin(30)t-\frac{1}{2}(9.8)t^{2}$

Solving for t we find t=2.04 seconds to reach that point.

That means it's x coordinate is $\displaystyle 20cos(30)(2.04)=35.35 \;\ m$

The stone is at (35.35,0) at 2.04 seconds.

Now, we have a triangle with angle 45 and side 35.35

So, the y component is the same, y=-35.35 feet

Plug this into the y component and solve for t.

$\displaystyle -35.35=20sin(30)t-\frac{1}{2}(9.8)t^{2}$

And we find t=3.89 seconds

Perhaps TopsQuark will check me out.

3. Hello, PFX!

A stone is projected from a point $\displaystyle O$ on a cliff with a speed of 20 m/s
at an angle of elevation of 30°.
t seconds later the angle of depression of the stone from $\displaystyle O$ is 45°.
Find the value of $\displaystyle t$.

The parametric equations for the position of the stone (relative to $\displaystyle O$) are:

. . $\displaystyle \begin{array}{ccccccc}x &=& (20\cos30^o)t & \Rightarrow & x &=& 10\sqrt{3} t \\ y &=& (20\sin30^o)t - 4.9t^2 & \Rightarrow & y &=& 10t - 4.9t^2 \end{array}$

Consider the position of the stone at time $\displaystyle t.$
Code:
         *
*   *
O* - - - +A
\    *:
\   :
\ :
*B

Since $\displaystyle \angle AOB = 45^o,\:\Delta OAB$ is an isosceles right triangle.
. . Hence: .$\displaystyle OA = AB$

This means: .$\displaystyle x \,=\,-y\quad\Rightarrow\quad (10\sqrt{3})t \;=\;-\left[10t - 4.9t^2\right]$

. . which simplfies to: .$\displaystyle 4.9t^2 - 10(1+\sqrt{3})t \:=\:0$

. . which factors: .$\displaystyle t\left[4.9t - 10(1+\sqrt{3})\right] \:=\:0$

. . and has roots: .$\displaystyle t = 0,\;\frac{10(1+\sqrt{3})}{4.9}$

Therefore: .$\displaystyle t \:\approx\:5.6$ seconds.

4. My thinking was on line with yours, Soroban. Why is my answer different?.

I don't see anything wring with my logic. The range of the stone is $\displaystyle \frac{400sin(60)}{9.8}=35.33$.

5. hi, thanx for ur replies....

well the correct answer given is T = 5.59 (to 3 s.f.)

@ galactus i cant find where u went wrong....

6. Originally Posted by galactus
I am including a diagram. I hope I am interpreting correctly.

We can use the horizontal and vertical component formulas.

horizontal: $\displaystyle x=(20cos(30))t$

vertical: $\displaystyle y=(20sin(30))t-\frac{1}{2}(9.8)t^{2}$

When the rock is level with the point it was thrown, then it's the point where y equals 0, we can find x at that time.

$\displaystyle 0=20sin(30)t-\frac{1}{2}(9.8)t^{2}$

Solving for t we find t=2.04 seconds to reach that point.

That means it's x coordinate is $\displaystyle 20cos(30)(2.04)=35.35 \;\ m$

The stone is at (35.35,0) at 2.04 seconds.

Now, we have a triangle with angle 45 and side 35.35

So, the y component is the same, y=-35.35 feet

Plug this into the y component and solve for t.

$\displaystyle -35.35=20sin(30)t-\frac{1}{2}(9.8)t^{2}$

And we find t=3.89 seconds

Perhaps TopsQuark will check me out.
Take a look at your diagram. When y = 0 you have x at 35.35 ft. How can it be that when it drops to -45 degrees? x is always increasing.

-Dan