Physics: Non-horizontal projectile motion?

Question

Howdy, everyone.

I'm solving a ballistic problem about a bi-dimensional motion of a projectile, thrown from coordinates x=0 and y=0.

Take that I've a degree a = 20° and an initial velocity Vo = 20.0 m/s. I'd solve the problem this way:
first of all, I'd find the x and y components of Vo, that would be VoX (20.0 m/s * cos a) and VoY (20.0 m/s * sin a).
Then onto the half-height time: Ti = Vfy - VoY / Ay (remembering that Vfy in this case is 0.0 m/s since the velocity in the peak of the motion is = 0.0 m/s).
Then I'd multiply this result by 2 to get the full time of the motion and get the total time Tf.
That way I can calculate the range: dx= VoX * tf.
To get the hmax (the peak height) I'd do: hmax= VoY * Ti + 1/2 g * t^2.
I wanted to know if this procedure is correct.

Then, another question. Put that I've the same problem (a = 20° and V0 = 20.0 m/s). But the projectile is thrown from a mountain (for istance) and the height of this cliff is 10 m.

How can I solve it?

Answer 1

Your 1/2 height equation is wrong. Note the units:
sec = m/sec - sec

Start with Vfy = Voy +Ay*Ti
Ti = (Vfy - Voy)/Ay

(Oh, I don't think I was reading your mind accurately, maybe that's what you meant.)

Then if t=Ti in your hmax equation, the rest of what you said looks OK.

Another question:
Use the same method to find hmax (above the top of the cliff). Then revise the hmax formula
hmax + 10m= VoY * Tj + 1/2 g * Tj^2
where Tj is the time from the peak to the valley floor. And then continue using Ti+Tj to find dx.

Answer 2

the two balls will hit the floor on the comparable time inspite of ways briskly interior the horizontal the two one is shifting. The acceleration on the two balls interior the vertical course is the comparable, so they're going to fall on the comparable value. notwithstanding in case you throw one horizontally, it is going to nonetheless hit on the comparable time because of fact the only that's not thrown horzontally. you may manage horizontal and vertical action seperatly.