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Old Jul 10th 2017, 04:32 PM   #1
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Object thrown off a balcony

Hi,

I've been here a lot lately, and I appreciate everybody's help. My lesson text has various errors in this lesson, and a lot of it is very unclear. It is really affecting how well I am learning the material, and it's hard to tell when I am correct or incorrect. Anyways, I'll stop complaining...

The question is:

An object is thrown upwards off a balcony at 2.1m/s, the acceleration is 9.81m/s^2 due to gravity, and it takes 3 seconds to hit the ground. How high is the balcony and how fast was it going when it hit the ground?

I've tried it with several of the equations that they introduced (but failed to explain) and the answer I believe to be correct is this, please let me know how I did:

V2 = V1 + at
V2 = 2.1m/s -9.81m/s^2 (3s)
V2 = -27.33m/s

It hit the ground at -27.33m/s.

d = 1/2(V1 + V2)t
d = 1/2(2.1m/s -27.33m/s)(3s)
d = -37.845m

The balcony is 37.845m high.

I got the same answer with d = V1(t) + 1/2(a)(t^2)

It feels right, but with this current lesson, I couldn't be any more unsure.

Thanks very much!
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Old Jul 10th 2017, 08:16 PM   #2
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height of the balcony is $|\Delta y|$ ...

$\Delta y = v_0 \cdot t - \dfrac{1}{2}gt^2$

$\Delta y = 2.1 \cdot 3 - \dfrac{1}{2}(9.81)(3^2) = -37.845$

$h = |-37.845| = 37.845 \, m$
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acceleration, balcony, displacement, initial, object, thrown, time, velocity



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