cq_1_082

phy201

Your 'cq_1_08.2' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.

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8.2

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?

answer/question/discussion: ->->->->->->->->->->->-> :

'ds= 12meters

V0= 15m/s

a =10m/s^2

Vf^2 = V0^2 + 2a * 'ds

Vf^2= (15m/s)^2 + 2(10m/s^2) * 12meters

Vf^2 = 225m^2/s^2 + 20m/s^2 * 12 meters

Vf^2= 225m^s/s^2 + 240m^2/s^2

Vf^2= 465m^2/s^2

sqrt vf^2= sqrt 465m^2/s^2

vf= 22m/s

Vf= v0 + a * 'dt

22m/s = 15m/s + 10m/s^2 * 'dt

7m/s = 10m/s^2 *'dt

'dt = .7 seconds

'ds = vave * 'dt

Vave= (15m/s + 22m/s) / 2

vave= 18.5m/s

'ds = 18.5m/s * .7sec

'ds = 13 meters

After .7 of a second the ball has gone 13 meters

#$&*

How fast is it then going when it hits the ground, and how long after the initial toss does it first strike

the ground?

answer/question/discussion: ->->->->->->->->->->->-> :

'ds = 13 meters (total to hit ground)

V0 = 15m/s

a = 10 m/s^2

I think we need vf

vf^2 = V0^2 + 2a * 'ds

Vf^2 = 15m/s + 2(10m/s) * 13meters

vf^2= 15m^2/s^2 + 20m/s * 13meters

Vf^2 = 15m^2/s^2 + 260m^2/s^2

Sqrt Vf^2= sqrt 275m^2/s^2

vf= 17m/s

vave= 16m/s

'ds = vave * 'dt

13meters = 16m/s * 'dt

'dt = .8 seconds

So if the ball start at an initial velocity of 15m/s and has gone up a total of 13 meters when it travels

back down the 13 meters at an acceleration consistent with gravity pulling at 10m/s^2 will be at a speed of

17m/s for .8 of a second.

#$&*

At what clock time(s) will the speed of the ball be 5 meters / second?

answer/question/discussion: ->->->->->->->->->->->-> :

Vf= 5 m/s

a= 10m/s^2

ds= 13 meters

Vf^2 = V0^2 + 2a * 'ds

(5m/s)^2= V0^2 + 2(10m/s^2) * 13meters

25m^/s^2 = V0^2 + 20 m/s^2 *13m

25m^2/s^2 = V0^2 + 260m^2/s^2

V0^2 = 235m^2/s^2

V0 = 15m/s

'ds = vave * 'dt

13m = 10m/s * 'dt

'dt = 1.3 seconds

#$&*

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: ->->->->->->->->->->->-> :

'ds = 20 meters

V0 = 15m/s

a = 10m/s^2

Vf^2 = V0^2 + 2a * 'ds

Vf^2 = (15m/s)^2 + 2(10m/s^2) * 20meters

Vf^2 = 225m^2/s^2 + 400m^2/s^2

Vf= 25m/s

vave= (25m/s + 15m/s) / 2

vave= 20m/s

ds= vave * 'dt

20meters = 20m/s * dt

dt= 1 second

'ds= vave * 6seconds

'ds = 20m/s * 6 seconds

'ds = 120 meters

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about an 30 min to an hour

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You're doing really well identifying quantities and using the equations. All you really need to do to get these problems is choose your positive direction, and give the various quantities + or - signs accordingly.

&#Please compare your solutions with the expanded discussion at the link

Solution

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