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The problem: (I'm having trouble with (ii) but I listed (i) because one of the answers depend on it)

Drawing it out would help.

The point O is 20m above horizontal ground. A particle is projected from O with velocity 35 m/s at an angle of elevation 45 degrees above the horizontal, and it moves freely under gravity. The particle hits the ground at the point A. Find:


(i) the height above the ground of the highest point of the path of the particle.

$v=u+at$ ... $u=35sin(45)$ ... $v=0$ ... $a=-10$ (gravitational force)

$0=35sin(45)-10t$ ... $t~= 2.47$

$s=ut+0.5at^2$ ... $s=35sin(45)*2.47-5(2.47)^2$

$s=30.6+20~=50.6$

(we added +20 in the end since O is 20 meters above the ground).

...Correct. The answer key says that this is the correct answer.

(ii) the time taken for the particle to travel from O to A.

okay

1st method of solution:

I considered the particle's motion from "the highest point" to A, got out the time and then added 2.47s to the answer

$s=ut+0.5at^2$ ... $s=-50.6$ ... $a=-10$ ... $u=0$

$-50.6=-5t^2$ $t=\sqrt(50.7/5)$ $t~=3.18$

$3.18+2.47=5.66s$

...Correct. The answer key says that this is the correct answer.

2nd method of solution:

I considered the particle's motion from O till A.

$s=ut+0.5at^2$ ... s=-30.6 ... $u=35sin(45)$ ... $a=-10$

$-30.6=35sin(45)t-5t^2$ ... $5t^2-35sin(45)t-30.6=0$

That's a quadratic equation! After solving it with the quadratic formula (I used the calculator's "quadratic formula solver") I get the positive answer to be:

$t=5.97$s

There are 0.3 seconds extra.

Where's my mistake? Maybe it's an accuracy issue? (due to approximations and so) What did I do wrong?

p.s: sometimes the answer key has mistakes, don't count on it much Thanks in advance :)

edit: The mistake was $s=-30.6$. The correct displacement value is $s=-20$ that's all! :D

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Pls format your answer using the math code! I just tried it but it's a bit hard –  mjb4 Feb 27 at 13:53
    
I would if I could. This is my first question on this site. I'll try :) –  Mina Michael Feb 27 at 13:55
    
thx, if you need some help I can edit the first formula so you know how to do the rest –  mjb4 Feb 27 at 13:58
    
I'm formatting it right now. seems easy :D just hang on a minute –  Mina Michael Feb 27 at 14:01
2  
How d'you get $s=-30.6$ the second way? –  mjb4 Feb 27 at 14:13

1 Answer 1

up vote 0 down vote accepted

Ok the particle rises for $t_{r} = 2.47 s$, then it falls again for the same time.
But hang on it is still $20m$ above ground. So we have to calculate how long it will take for the rest.
The Velocity $v_{f}$ for the last $20m$ is exactly the velocity the particle had as you were shooting it upwards (remember this is a parabola)!
So for the movement you will get:

$$ h(t) = 20m-35\sin(45°)t-\frac{10}{2}t^2 = 0 \quad\Rightarrow \quad t^2+\frac{2\cdot 24.75}{10}t-20 =0 $$

Solving this gives you one rubish solution (it's negative and one positiv, thats the one!): $$ t_1 = -\frac{35\sin(45°)}{2}+\sqrt{\left(\frac{35\sin(45°)}{2}\right)^2+20} = 0.7833 s $$

So the total flight time is: $$ T = 2t_r+t_1 = 5.723 $$

You're sure are right with 5.66 seconds

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yes I'm definitely sure that the correct time answer is 5.66. I'm looking over things now. –  Mina Michael Feb 27 at 15:06
    
Hey! You're considering different points for the ball's motion. This means that this is a 3rd method of answering. (it doesn't relate to the 2 methods I'm asking about up there) ... anyways let's solve it your way. Now your assumption $T = 2t_r+t_1$ seems correct to me. But I can't identify the 2 equations you're using to determine $t_1$ –  Mina Michael Feb 27 at 15:17
    
$s=s+ut-0.5at^2=0$ ???? I think this equation is messed up –  Mina Michael Feb 27 at 15:19
    
Ok the particle is shot up in the air it travels for $t_r$ seconds, then it will fall again for $t_r$ seconds. But in this fall it will gain the velocity, it had when it was shot up in the air $u$, just negative now! So it will fall $h=20m$ from that point onwards giving you $s(t) = h-ut-0.5at^2$ and this is supposed to be zero because at height zero it landed! –  mjb4 Feb 27 at 15:22
    
well I understand all you're saying and I agree but it's just that equation that doesn't fit in my head!! if you try replacing this equation $s(t)=h−ut−0.5at^2$ with this: $s=ut+0.5at^2$ it would give you $-20=-35sin(45)t-5t^2$ which would give you the quadratic equation $5t^2+35sin(45)t-20=0$ whose positive answer is 0.707 ... input 0.707 to the equation $T = 2t_r+t_1$ and you get 5.66 seconds –  Mina Michael Feb 27 at 15:29

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