Some Basic Physics

For some reason, I can't get a right answer despite the relative simplicity of this problem:

A launcher fires a ball 3.16 m/s at an angle of 26 degrees from horizontal.  If the launcher fired the ball from a height of 127.3 cm., how far away from the launcher would the ball land?  (Assume negligible air resistance, wind, etc.)

What's more, I need to do the same thing for an initial velocity of 4.98 and 6.84, and show my work.  I've tried various different approaches, and all of them either came up with different answers or answers that didn't make any sense.  I'm usually good at physics, but for some reason I just can't do this one.  Can I get some help on this one?

lern to highschooled

There is a vertical component to the motion and a horizontal component to the motion (well mathematically we say this).  First you resolve the given velocity vector v into the two components vx and vy.  So vx = v times cos(angle) and vy = v times sin(angle).  Those formulas are from basic trigonometry.  Anyway by ignoring air resistance you get to assume that horizontal vx stays constant throughout the whole trip.  The vertical vy however is subject to change because of the force of gravity.

OK I'm tired maybe I'll continue soon.

And be sure you converted 127.3 cm into metres because your velocity units are m/s.

So anyway you could solve for the time it takes for vy to slow down to zero as the ball gets at its maximum height, then you could solve for far vertically the ball went to get up there, then you could add that distance to the distance off the ground it started from, then you could solve for the time it takes for the ball to fall down from that total distance, then you could add the two different times that you have solved (the up time and the down time).  So you would know the total time in motion.

Formula vy = initial vy - g times t, but rearranged to solve for time: t = (initial vy - vy)/g.  To get time to maximum height, realize that vy would be zero at that time.

To get distance travelled up, formula: dy = initial vy times t - 1/2 times g times t^2.

To get time to fall total distance down, formula is previous fomula solving for t, set initial vy at zero now because the desired start is at the maximum height when the ball is hanging there in space.  t = (square root of (2 times g times dy))/g

Add time to go up plus time to go down for total time in air.

Then you could use that time knowledge in the horizontal case to calculate how far sideways the ball got, with simply d = vx times t.

Could be errors on my part here, at least it was free help.

http://www.ajdesigner.com/phpprojectilemotion/vertical_velocity_equation.php

Ah, I think I've got it now.  My mistake was not separating out the going up part of the arc from the going down part.  Thanks.

Yay!

Come to think of it, it could be a fine alternative to use a single formula to find time to achieve velocity vy equal to but opposite in direction from the initial upward vy, then find time to travel the extra 127.3 cm down using that vy as intial velocity, then add those two times together to get total time.

That's the best way to do it if you're man fires it from ground height, yes. Even simpler in this case is to use
s = ut + 1/2at^2 in the vertical direction, substituting in -1.273 for s, 3.16 for u and g for a, to get a quadratic in t, then solve and take the positive solution to get the total time.

use calculus.

42.  THE ANSWER IS 42.

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Marijuana MUST be legalized.