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A ball is tossed upward at 15 meters / second from a height of 12 meters above the ground. Assume a uniform downward acceleration of 10 m/s^2 (an approxomation within 2% of the 9.8 m/s^2 acceleration of gravity).
How high does it rise and how long does it take to get to its highest point?
V0=15
Vf=0
a=-10
vf=vo +at
0=15 +-10t
t=1.5seconds
`ds = v0 `dt + .5 a `dt^2
'ds=15 1.5 +.5 -10 1.5^2
'ds=11.25m
How fast is it then going when it hits the ground, and how long after the initial toss does it first strike the ground?
'ds=11.25+12=23.25m
Vo=0
a=-10m/sec^2
vf^2 = v0^2 + 2 a `ds.
=0 + 2 -10 23.25m
vf=21.56m/sec
vf = v0 + a * `dt
21.56=0 +-10 x dt
dt=2.15 seconds
This result gives the number of seconds required to fall from the highest point to the ground. This will occur 1.5 sec + 2.15 sec = 3.65 sec after the specified initial instant.
It isn't necessary to solve for the maximum height (which in this problem was previously requested) in order to find the result requested here. You can start with the given initial information and measure `ds with respect to the initial point. The analysis is as follows:
v0 = 15 m/s
`ds = -12 m
a = -10 m/s^2
Solving vf^2 = v0^2 + 2 a `ds for vf we get
vf = +- sqrt( v0^2 + 2 a `ds)
= +- sqrt( (15 m/s)^2 + 2 * (-10 m/s^2) * (-12 m) )
= +- sqrt( 225 m^2/s^2 + 240 m^2/s^2)
= +- sqrt(465 m^2/s^2)
= +-21.6 m/s, approximately.
From the specified conditions we know that the + solution is not reasonable, so we have
vf = - 21.6 m/s.
With this and the three initial quantities we can easily solve for or reason out `dt.
The simplest reasoning is that vAve = (+15 m/s + (-21.6 m/s) ) / 2 = -3.3 m/s, so the time interval is `dt = `ds / vAve = -12 m / (3.3 m/s) = 3.6 sec, approximately.
Another way to solve directly for `dt is to use the same information in the third equation. This solution is a little messy and if your algebra skills aren't in good shape, you might want to skip it:
`ds = v0 `dt + .5 a `dt^2 is quadratic in `dt. Putting the equation into the general form of a quadratic we get
.5 a `dt^2 + v0 `dt - `ds = 0.
This is of the form A x^2 + B x + C = 0, the standard form of a quadratic equation, where x is used in place of `dt and A = .5 a, B = v0 and C = -`ds. The solution is
x = (-B +- sqrt( B^2 - 4 A C ) ) / (2 A)
= (-v0 +- sqrt( v0^2 - 4 * (.5 a) * (-`ds) ) / (2 ( .5 a) )
= (-v0 +- sqrt( v0^2 + 2 a `ds) ) / a.
Since x was used to stand for `dt we have
`dt = (-v0 +- sqrt( v0^2 + 2 a `ds) ) / a
= (-15 m/s +- sqrt( (15 m/s)^2 + 2 * 10 m/s^2 * (-12 m) ) ) / (-10 m/s^2)
= (-15 m/s +- sqrt( 465 m^2 / s^2) ) / (-10 m/s^2)
= (-15 m/s +- 21.6 m/s) / (-10 m/s^2)
= -36.6 m/s / (-10 m/s^2) OR (+6.6 m/s) / (-10 m/s^2)
= 3.6 s OR -0.66 s.
The 3.6 s solution is consistent with the preceding solution for `dt.
The -0.66 s solution indicates that if the projectile was already in uniformly accelerated motion prior to the 'initial' instant, it passed through the -12 m position 0.66 s before passing through the 'initial' instant.
How high will it be at the end of the sixth second?
ds = v0 `dt + .5 a `dt^2
ds = 0 (6-1.5)+.5 -10 4.5^2
'ds=101m
That would be -101 m. It would be 101 m below its initial position. However this assumes that v0 = 0, meaning that you would be starting the clock when the object is at its maximum height,position would be measured relative to that height and time interval is reduced by the 1.5 seconds required to rise to the maximum height.
Using v0 = +15 m/s and using the full 6 seconds we get
`ds = 15 m/s * 6 s + .5 * (-10 m/s^2) * (6 s)^2 = 90 m - 180 m = -90 m.
This is the displacement relative to the initial position. Relative to the max-height position this would be -90 m - 11.25 m = -101.25 m, in agreement with your calculation.
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Good work overall, but be sure to see my notes for some more (and in some cases less) efficient ways to set parts of this problem.