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# maths mechanics M1 SUVAT Q help watch

1. hey, can anyone help me with this question

A stone is projected vertically upwards from ground level with a speed of 20ms-1, 1 second later a second stone is projected vertically upwards from the same point with a speed of 25ms-1. Find:
a) how long after the first stone is release do they collide
b) the height at which the collide
2. (Original post by hanaxogrunge)
hey, can anyone help me with this question

A stone is projected vertically upwards from ground level with a speed of 20ms-1, 1 second later a second stone is projected vertically upwards from the same point with a speed of 25ms-1. Find:
a) how long after the first stone is release do they collide
b) the height at which the collide
What have you tried? Try forming one equation for each stone, then try to combine them using the fact that s (the height above the ground) will be the same for both stones when they collide.

If in doubt, write down the variables you know (in this case, u and a) and those you're trying to find out (s and t). Which equation involves these four variables?
3. (Original post by TheMindGarage)
What have you tried? Try forming one equation for each stone, then try to combine them using the fact that s (the height above the ground) will be the same for both stones when they collide.

If in doubt, write down the variables you know (in this case, u and a) and those you're trying to find out (s and t). Which equation involves these four variables?
i did kinda try that but ended up confusing myself

the equation which includes all of those is u+v/2*t = s
so do i plug those numbers in
but then what do i do with t and s
4. (Original post by hanaxogrunge)
i did kinda try that but ended up confusing myself

the equation which includes all of those is u+v/2*t = s
so do i plug those numbers in
but then what do i do with t and s
You don't want v in your equation - you want a (since a is just the acceleration due to gravity). Try s = ut + 0.5at^2.

Plug in the values of u for each stone. For the first stone, leave t as it is, but for the second, use (t-1) because it is projected one second later. Now, when the stones collide, the two values of s are equal, so you can set both of the right-hand-sides of the equation equal to each other.
5. (Original post by TheMindGarage)
You don't want v in your equation - you want a (since a is just the acceleration due to gravity). Try s = ut + 0.5at^2.

Plug in the values of u for each stone. For the first stone, leave t as it is, but for the second, use (t-1) because it is projected one second later. Now, when the stones collide, the two values of s are equal, so you can set both of the right-hand-sides of the equation equal to each other.
ahhhh
okay

thank you so much
6. a is going to be -9.8 because it is projected vertically upwards so against gravity

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