Phy 231
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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 approximation 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?
How fast is it then going when it hits the ground, and how long after the initial toss does it first strike the ground?
At what clock time(s) will the speed of the ball be 5 meters / second?
At what clock time(s) will the ball be 20 meters above the ground?
How high will it be at the end of the sixth second?
answer/question/discussion:
To reach its highest point takes:
'dt = (vf - v0) / a
'dt = (0 m/s - 15 m/s) / -10 m/s^2 = 1.5 s
The highest point is found by:
'ds = (vf + v0) / 2 * 'dt
'ds = (0 m/s + 15 m/s) / 2 * 1.5 s = 7.5 m/s * 1.5 s = 11.25 m
adding 11.25 m to the starting position of 12 m, gives: 23.25 m
When it hits the ground its speed is:
When the ball strikes the ground, the 'ds = -12 m, therefore:
vf^2 = v0^2 + 2 'ds a
vf^2 = (15 m/s)^2 + 2 * -12 m * -10 m/s^2
vf^2 = 225 m^2/s^2 + 240 m^2/s^2
sqrt(vf^2) = sqrt(465 m^2/s^2)
vf = +- 21.56 m/s (22 m/s, rounded)
Since we know it's travelled downward, it would be -21.56 m/s
To strike the ground, it takes:
'dt = 'ds / vAve
'dt = -12 m / [(vf + v0) / 2]
'dt = -12 m / [(-21.56 m/s + 15 m/s) / 2]
'dt = -12 m / -3.28 m/s = 3.66 s
It will take ___ s to reach 5 m/s:
vf = 5 m/s, v0 = 15 m/s, a = -10 m/s^2
'dt = (vf - v0) / a
'dt = (5 m/s - 15 m/s) / -10 m/s^2 = -10 m/s / -10 m/s^2 = 1 s
It will take ____ s to reach 20 m:
'ds = 8 m, v0 = 15 m/s, a = -10 m/s^2
vf^2 = v0^2 + 2 a 'ds
vf^2 = (15 m/s)^2 + 2 * -10 m/s^2 * 8 m
vf^2 = 225 m^2/s^2 -160 m^2/s^2
sqrt(vf^2) = sqrt(60 m^2/s^2)
vf = +- 8.06 m/s
Because the ball is still moving upward, the velocity will be positive: 8.06 m/s
'dt = 'ds / [(vf + v0) / 2]
'dt = 8 m / [ (8.06 m/s + 15 m/s) / 2 ]
'dt = 8m / 11.53 m/s
'dt = 0.69 s
After 6 seconds it will be:
'dt = 6 s, v0 = 15 m/s, a = -10 m/s^2
a = (vf - v0) / 'dt
vf = a * 'dt + v0 = -10 m/s^2 * 6s + 15 m/s = -60 m/s + 15 m/s = -45 m/s
'ds = (vf + v0) / 2 * 'dt
'ds = (-45 m/s + 15 m/s) / 2 * 6s = -30 m/s / 2 * 6s = -90 m
Adding the original starting point of +12 m, we arrive at: -78 m
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30 mins
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Excellent work.