[SOLVED] Projectiles....

**PFX** · May 14th 2008, 11:55 AM

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...

**galactus** · May 14th 2008, 01:10 PM

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

We can use the horizontal and vertical component formulas.

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

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

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.

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

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

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

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.

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

And we find t=3.89 seconds

Perhaps TopsQuark will check me out.

**Soroban** · May 14th 2008, 01:22 PM

Hello, PFX!

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°.
t seconds later the angle of depression of the stone from $O$ is 45°.
Find the value of $t$ .

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

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

Consider the position of the stone at time $t.$

Code:

         *
       *   *
     O* - - - +A
        \    *:
          \   :
            \ :
              *B

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

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

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

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

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

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

**galactus** · May 14th 2008, 02:17 PM

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 $\frac{400sin(60)}{9.8}=35.33$ .

**PFX** · May 14th 2008, 05:39 PM

hi, thanx for ur replies....

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

so Soroban's answer is correct

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

**topsquark** · May 14th 2008, 06:18 PM

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: $x=(20cos(30))t$

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

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.

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

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

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

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.

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

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

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