http://imageshack.us/photo/photo/850/ima…
I took a picture of my calculations. The question asked to find delta x of the problem. Meaning how far the ball lands. Could you see if I did anything wrong?
I took a picture of my calculations. The question asked to find delta x of the problem. Meaning how far the ball lands. Could you see if I did anything wrong?
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The method is OK but...
The initial vertical component of velocity, VIy, is negative as the ball is moving downwards.
Δy = -45.0m (you have written this without the minus sign, but used the correct value in the formula).
So you should have (omitting units):
-45.0 = -1.06t + ½(-9.81)t²
4.905t² + 1.06t - 45.0 = 0
Which I think gives t = 2.923s
It is worth working to at least 4 significant figures during the calculation and only rounding to 3 at the very end. This minimises rounding errors during the calculation.
The initial vertical component of velocity, VIy, is negative as the ball is moving downwards.
Δy = -45.0m (you have written this without the minus sign, but used the correct value in the formula).
So you should have (omitting units):
-45.0 = -1.06t + ½(-9.81)t²
4.905t² + 1.06t - 45.0 = 0
Which I think gives t = 2.923s
It is worth working to at least 4 significant figures during the calculation and only rounding to 3 at the very end. This minimises rounding errors during the calculation.
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Horizontal component = (cos 45) x 1.5, = 1.061m/sec.
Initial vertical component = (sin 45) x 1.5, = 1.061m/sec.
Vf^2 = Vi^2 + 2gh.
(1.061^2 + 2(9.8 x 45) = Vf^2, = 883.125721.
Vf = 29.72m/sec. Subtract Vi, = 28.659m/sec gain.
Time = (v/g) = 2.92439 secs.
(2.92439 x 1.061) = 3.103 metres.
Initial vertical component = (sin 45) x 1.5, = 1.061m/sec.
Vf^2 = Vi^2 + 2gh.
(1.061^2 + 2(9.8 x 45) = Vf^2, = 883.125721.
Vf = 29.72m/sec. Subtract Vi, = 28.659m/sec gain.
Time = (v/g) = 2.92439 secs.
(2.92439 x 1.061) = 3.103 metres.