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
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