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A level Maths

At time T=0, particle A is dropped from a bridge 40m above the ground. One second later particle B is projected vertically upwards with speed 5ms-1 from a point 10m above the ground.
A) find the distance travelled by A when B is at the highest point in motion.
B) how long after A is dropped do the two particles become level?
C) How far from the ground are they at this time
Reply 1
Avatar for mqb2766
mqb2766
21
Original post by malli07
At time T=0, particle A is dropped from a bridge 40m above the ground. One second later particle B is projected vertically upwards with speed 5ms-1 from a point 10m above the ground.
A) find the distance travelled by A when B is at the highest point in motion.
B) how long after A is dropped do the two particles become level?
C) How far from the ground are they at this time


What have you tried / what are you stuck with?
Reply 2
Avatar for Squirmz
Squirmz
17
For A) Use “v=u+at” to find the time taken for B to reach max height (u=5, a=-g, v=0). Then you say that A has been travelling for 1 second more than B, so sub t+1 into the equation “s=ut+1/2 at^2” to (a=-g, u=0, t=t1+1). The value for s will probably come out negative so just explain that it’s a distance which is scalar and write a positive answer in conclusion.

For b) You’ll need to set up simultaneous equations in terms of t for the displacements of each particle, ensuring that you use the same positive direction. Then solve for t. You’ll have to let the height of the particles when they are level be a variable (x or h), and deal with the starting heights.

c) Take your answer to b) to find h.

Hope this helped, I can explain b) in a bit more detail of you need 😃
Reply 3
Avatar for malli07
malli07
OP
8
Original post by mazenod
For A) Use “v=u+at” to find the time taken for B to reach max height (u=5, a=-g, v=0). Then you say that A has been travelling for 1 second more than B, so sub t+1 into the equation “s=ut+1/2 at^2” to (a=-g, u=0, t=t1+1). The value for s will probably come out negative so just explain that it’s a distance which is scalar and write a positive answer in conclusion.

For b) You’ll need to set up simultaneous equations in terms of t for the displacements of each particle, ensuring that you use the same positive direction. Then solve for t. You’ll have to let the height of the particles when they are level be a variable (x or h), and deal with the starting heights.

c) Take your answer to b) to find h.

Hope this helped, I can explain b) in a bit more detail of you need 😃


Ok so I understood part A.... can you please explain part b?

Thank you 😊
Reply 4
Avatar for mqb2766
mqb2766
21
Original post by malli07
Ok so I understood part A.... can you please explain part b?

Thank you 😊

Try setting up displacement equations for A and B. When the particles meet, they will have the same displacement at the same time so you can equate them to solve for t.
Reply 5
Avatar for Squirmz
Squirmz
17
Original post by malli07
Ok so I understood part A.... can you please explain part b?

Thank you 😊


So for b, you write the displacements of the particles in terms of the height that they are at when level, so for particle A, with up as the positive direction, the displacement of the particle will be s=h-40, and for b, the displacement with up as the positive direction is s=h-10. So you can use “s=ut 1/2 at^2” on each one. You know acceleration, and you know the initial velocities, so set up two simultaneous equations in terms of h and t, and set one equal to the negative of the other (since the displacements are in opposite directions) . 🙂 Remember for particle B, you should sub time as (t-1) since it begins its’ motion 1 second later than A.
(edited 3 years ago)

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