Can someone help with this maths questions on the SUVAT equations?
A particle P1 is projected vertically upwards, from horizontal ground, with a speed of 30 m s-1. At the same instant another particle P2 is projected vertically upwards from the top of a tower of height 25 m, with a speed of 10 m s-1. Find.
(i)The time for which P1 is higher than the top of the tower,
(ii)The velocities of the particles at the instant when the particles are at the same height,
(iii)The time for which P1 is higher than P2 and is moving upwards.
(i)The time for which P1 is higher than the top of the tower,
(ii)The velocities of the particles at the instant when the particles are at the same height,
(iii)The time for which P1 is higher than P2 and is moving upwards.
(edited 1 year ago)
ok i just did i)
i this case: s=25 (because we want to find when it hits 25 on the way up and on the way down), u=30, a=-9.8 (because it’s acceleration due to gravity and we’re going against the gravitational force), t=?
so we use s=ut+0.5at^2
and we get a quadratic: 4.9t^2-30t+25=0
solve that and you get t=0.995 (when it hits 25 on the way up) and t=5.13 (when it hits 25 on the way down).
so we just subtract the two and you get t=4.14s (3sf)
i this case: s=25 (because we want to find when it hits 25 on the way up and on the way down), u=30, a=-9.8 (because it’s acceleration due to gravity and we’re going against the gravitational force), t=?
so we use s=ut+0.5at^2
and we get a quadratic: 4.9t^2-30t+25=0
solve that and you get t=0.995 (when it hits 25 on the way up) and t=5.13 (when it hits 25 on the way down).
so we just subtract the two and you get t=4.14s (3sf)
What have you tried so far?
Usually with mechanics questions, your first two tasks is to:
(i) draw a nice diagram; and
(ii) write down what is given in the question, so for instance u=?, s=?, t="to be found", etc.
Usually with mechanics questions, your first two tasks is to:
(i) draw a nice diagram; and
(ii) write down what is given in the question, so for instance u=?, s=?, t="to be found", etc.
Original post by abimt78
...
Thanks for the help, but note that we don't do full solutions here.
Original post by tonyiptony
Thanks for the help, but note that we don't do full solutions here.
oh sorry i didn’t realise
Original post by abimt78
ok i just did i)
i this case: s=25 (because we want to find when it hits 25 on the way up and on the way down), u=30, a=-9.8 (because it’s acceleration due to gravity and we’re going against the gravitational force), t=?
so we use s=ut+0.5at^2
and we get a quadratic: 4.9t^2-30t+25=0
solve that and you get t=0.995 (when it hits 25 on the way up) and t=5.13 (when it hits 25 on the way down).
so we just subtract the two and you get t=4.14s (3sf)
i this case: s=25 (because we want to find when it hits 25 on the way up and on the way down), u=30, a=-9.8 (because it’s acceleration due to gravity and we’re going against the gravitational force), t=?
so we use s=ut+0.5at^2
and we get a quadratic: 4.9t^2-30t+25=0
solve that and you get t=0.995 (when it hits 25 on the way up) and t=5.13 (when it hits 25 on the way down).
so we just subtract the two and you get t=4.14s (3sf)
Ok thanks i get the first one now, how did you do the next bit?
Original post by tonyiptony
What have you tried so far?
Usually with mechanics questions, your first two tasks is to:
(i) draw a nice diagram; and
(ii) write down what is given in the question, so for instance u=?, s=?, t="to be found", etc.
Usually with mechanics questions, your first two tasks is to:
(i) draw a nice diagram; and
(ii) write down what is given in the question, so for instance u=?, s=?, t="to be found", etc.
ive got the fist part of the question now I'm just stuck on the second
Original post by Grace xox
ive got the fist part of the question now I'm just stuck on the second
So I assume you have drawn a nice diagram, and stuff.
Okay, what does it mean mathematically that the two particles are "at the same height"?
(I haven't worked out yet, but suvat tend to just be plug-and-chug so if you solve this little mystery, you're good to go)
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