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Mechanics SUVAT question

A stone is fired vertically upwards from a catapult and lands 5.0s later. What was the initial velocity of the stone? For how long was the stone at a height of 20m or more?

My answer: 25m/s for part a
but the second answer is 3s and I don't know how. Is this a misprint?
Original post by Mad Man
A stone is fired vertically upwards from a catapult and lands 5.0s later. What was the initial velocity of the stone? For how long was the stone at a height of 20m or more?

My answer: 25m/s for part a
but the second answer is 3s and I don't know how. Is this a misprint?

Post your working please :smile:
Reply 2
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Mad Man
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s=0
u=?
v=v
a=-10m/s
t=5.0

use s=ut+1/2 at^2 then u =25m/s

hiwo do i do the second bit?

I did
s>20
u=25
a=-10

then formed a quadratic to solve t and got t=4 t=1
Original post by Mad Man
s=0
u=?
v=v
a=-10m/s
t=5.0

use s=ut+1/2 at^2 then u =25m/s

hiwo do i do the second bit?

I did
s>20
u=25
a=-10

then formed a quadratic to solve t and got t=4 t=1

Yes so at t= 1 it reaches 20 m and carries on upwards then stops and comes down reaching 20m after 4 seconds - so it is above 20 m for 4 - 1 = 3 seconds :smile:
Reply 4
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Mad Man
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please help
Reply 5
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Mad Man
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Original post by Muttley79
Yes so at t= 1 it reaches 20 m and carries on upwards then stops and comes down reaching 20m after 4 seconds - so it is above 20 m for 4 - 1 = 3 seconds :smile:

Oh thanks. I've never seen a question like this before. How would I know that? These suvat questions keep catching me out , I'm getting stressed.
Original post by Mad Man
Oh thanks. I've never seen a question like this before. How would I know that? These suvat questions keep catching me out , I'm getting stressed.

Look back at the question and think about the stone going up and down - you solved the quadratic and got two values. These are the times when the stone is exactly 20 m above the ground ... we want the time the stone is above 20 metres.

You need to 'see' scenarios in your head if you can :smile:
Reply 7
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Mad Man
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Original post by Muttley79
Look back at the question and think about the stone going up and down - you solved the quadratic and got two values. These are the times when the stone is exactly 20 m above the ground ... we want the time the stone is above 20 metres.

You need to 'see' scenarios in your head if you can :smile:

Note taken. :smile:
Reply 8
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Mad Man
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Original post by Muttley79
Look back at the question and think about the stone going up and down - you solved the quadratic and got two values. These are the times when the stone is exactly 20 m above the ground ... we want the time the stone is above 20 metres.

You need to 'see' scenarios in your head if you can :smile:

Can you explain this algebraically?
Original post by Mad Man
Can you explain this algebraically?

Draw a graph of height verses time. Mark the times that it above 20m. Does that make it clearer?
Original post by Mad Man
Can you explain this algebraically?


s=20
u=25
a=-10

Using s = ut + 0.5 t^2

20 = 25 t - 5t^2

5^2 - 25 t + 20 = 0
t^2 - 5t + 4 = 0

then (t - 4)(t - 1) = 0 so t = 1 or 4

we need the time period in between when 25 t - 5t^2 >20
Reply 11
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Mad Man
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Original post by RogerOxon
Draw a graph of height verses time. Mark the times that it above 20m. Does that make it clearer?

Yes it makes sense now. I guess I didn't read the question carefully.
Screenshot 2020-10-03 at 10.42.00 PM.png200.0KB
I’m not a physics expert but couldn’t you just use a = v-u / t and rearrange to t?

(My physics teacher says that if ut in s = ut + 1/2at^2 does not cancel then you should immediately stop as it leads to quadratic and choose another formula)

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