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# physic questions p2?

Exam coming up, prof did not provide solutions. Anything helps, prefer if you show approach.

Thanks,

### 2 Answers

- 2 months agoFavorite Answer
I'm going to use a bit of calculus for my own clarification. No need to pay attention just yet

a(t) = -9.8

v(t) = -9.8 * t + C

s(t) = -4.9 * t^2 + Ct + D

s(0) = 50.0

s(t) = 50 + C * t - 4.9 * t^2

Now you can pay attention.

The first stone is thrown downward with a velocity of 2.00 m/s, so C = -2 (it's negative because velocity is a vector and we're moving downward)

Let m(t) be the function for the first stone

m(t) = 50 - 2t - 4.9t^2

Find when m(t) = 0

0 = 50 - 2t - 4.9 * t^2

4.9 * t^2 + 2t - 50 = 0

49t^2 + 20t - 500 = 0

t = (-20 +/- sqrt(400 + 4 * 49 * 500)) / 98

t = (-20 +/- 2 * sqrt(100 + 49 * 500)) / 98

t = (-20 +/- 20 * sqrt(1 + 49 * 5)) / 98

t = -20 * (1 +/- sqrt(1 + 245)) / 98

t = -10 * (1 +/- sqrt(246)) / 49

It wouldn't make sense for t < 0

t = -10 * (1 - sqrt(246)) / 49

t = 10 * (sqrt(246) - 1) / 49

t = 2.9968137023179840682009503010742

We can say that t = 3.00

n(t) will be our next stone's path

n(t) = -4.9 * t^2 + Ct + 50

When t = 2, since 2 = 3 - 1, then n(t) = 0

0 = -4.9 * 2^2 + C * 2 + 50

4.9 * 4 - 50 = 2C

4.9 * 2 - 25 = C

9.8 - 25 = C

-15.2 = C

It's thrown down at 15.2 m/s

m'(t) = -2 - 9.8 * t

m'(3) = -2 - 9.8 * 3 = -2 - 29.4 = -31.4

n'(t) = -15.2 - 9.8 * t

n'(2) = -15.2 - 19.6 = -34.8

Look, I'm sure that the professor didn't give you solutions, but it's not like this is the first time you've seen problems like these. Draw some diagrams, look at your notes for projectile motion, and work it out, step by step. Sometimes, that's all you can do.

150 * cos(120) + m * cos(t) = 140 * cos(35)

150 * sin(120) + m * sin(t) = 140 * sin(35)

m * cos(t) = 10 * (14 * cos(35) - 15 * cos(120))

m * sin(t) = 10 * (14 * sin(35) - 15 * sin(120))

m^2 * (cos(t)^2 + sin(t)^2) = 10^2 * (196 * cos(35)^2 - 420 * cos(35) * cos(120) + 225 * cos(120)^2 + 196 * sin(35)^2 - 420 * sin(35) * sin(120) + 225 * sin(120)^2)

m^2 * 1 = 10^2 * (196 + 225 - 420 * cos(120 - 35))

m^2 = 100 * (421 - 420 * cos(85))

m = 10 * sqrt(421 - 420 * cos(85))

m = 196.05983475612325607405880556032

150 * cos(120) + 196 * cos(t) = 140 * cos(35)

196 * cos(t) = 140 * cos(35) - 150 * cos(120)

cos(t) = (140 * cos(35) - 150 * cos(120)) / 196

t = cos-1((140 * cos(35) - 150 * cos(120)) / 196)

t = +/- 14.588063297951140303009775328656

150 * sin(120) + 196 * sin(t) = 140 * sin(35)

sin(t) = (140 * sin(35) - 150 * sin(120)) / 196

sin(t) = (70 * sin(35) - 75 * sin(120)) / 98

t = sin-1((70 * sin(35) - 75 * sin(120)) / 98)

t = -14.659673478957222794754062525788

The discrepancy comes from my rounding of the magnitude before. The direction is at -14.6 degrees with respect to the x-axis.

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- Anonymous2 months ago
Nah, do your own homework

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