golgiapparatus31
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#1
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#1
https://imgur.com/a/lWPZm3w

For (a) I'm not getting the right answer of 0.77
(g is 10)

I tried the following:
Letting
t be the time to reach the window
u be the initial speed
theta be the angle of projection

I have
ucosθ t = 12
usinθ t - 5t^2 = 4
so usinθ = (4+5t^2)/t

Using the angle to the horizontal at the window
tan 30 = - Vy/Vx
tan 30 = -(usinθ - 10t)/(ucosθ)
ucos θ tan 30 = 10t - usinθ
using ucosθ t = 12 I get
12tan30 / t = 10t - usinθ
replacing usinθ = (4+5t^2)/t and solving gives t = 1.48

Please help
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Gundabad(good)
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#2
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#2
Maths is very hard. Just wanted to let you know.
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Smallwatch
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#3
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#3
For part a, only consider the vertical components. If you ignore the horizontal motion of the ball then a = g, and s = 4, and since the ball is starting from rest u = 0. You can then use a suvat equation to find a value for t. The second distance and angle are probably used in part b along with the answer from part a.
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ghostwalker
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#4
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#4
(Original post by golgiapparatus31)
https://imgur.com/a/lWPZm3w
Please help
'Fraid I'm unable to access imgur.com - at least until I buy a new computer.

If it's not already resolved, can you upload an image to TSR? Or someone else should be able to help.
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mqb2766
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#5
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#5
(Original post by golgiapparatus31)
https://imgur.com/a/lWPZm3w

For (a) I'm not getting the right answer of 0.77
(g is 10)

I tried the following:
Letting
t be the time to reach the window
u be the initial speed
theta be the angle of projection

I have
ucosθ t = 12
usinθ t - 5t^2 = 4
so usinθ = (4+5t^2)/t

Using the angle to the horizontal at the window
tan 30 = - Vy/Vx
tan 30 = -(usinθ - 10t)/(ucosθ)
ucos θ tan 30 = 10t - usinθ
using ucosθ t = 12 I get
12tan30 / t = 10t - usinθ
replacing usinθ = (4+5t^2)/t and solving gives t = 1.48

Please help
You get 0.77 from
t = sqrt(4(sqrt(3)-1)/5)
So it looks like a typo (wrong sign of -1) in the model answer? Yours looks good.
Last edited by mqb2766; 4 weeks ago
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golgiapparatus31
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#6
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(Original post by ghostwalker)
'Fraid I'm unable to access imgur.com - at least until I buy a new computer.

If it's not already resolved, can you upload an image to TSR? Or someone else should be able to help.
Thank you!!

(Original post by mqb2766)
You get 0.77 from
t = sqrt(4(sqrt(3)-1)/5)
So it looks like a typo (wrong sign of -1) in the model answer? Yours looks good.
Thank you!! Since answer for (b) is dependent on (a), can you please check my answer of 13.0 m/s for part (b) ? [12.9555431]
Replacing value of t, I get
ucostheta = 8.1169...
usintheta = 10.0975...
Using pythagoras gives u = sqrt(vx^2 + vy^2) = 13.0
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mqb2766
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(Original post by golgiapparatus31)
Thank you!!


Thank you!! Since answer for (b) is dependent on (a), can you please check my answer of 13.0 m/s for part (b) ? [12.9555431]
Replacing value of t, I get
ucostheta = 8.1169...
usintheta = 10.0975...
Using pythagoras gives u = sqrt(vx^2 + vy^2) = 13.0
From a quick check that looks good.
Tbh, a quick sanity check on the vertical velocity, you must have t>1 for it to be heading downwards with a velocity ~ -5 at a displacement of 4.
Last edited by mqb2766; 4 weeks ago
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golgiapparatus31
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#8
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(Original post by mqb2766)
From a quick check that looks good.
Tbh, a quick sanity check on the vertical velocity, you must have t>1 for it to be heading downwards with a velocity ~ -5 at a displacement of 4.
Thank you!
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