# Help with M1 question

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#1
At time t = 0, two balls A and B are projected vertically upwards. The ball A is projected vertically upwards with speed 2m/s from a point 50m above the horizontal ground. The ball B is projected vertically upwards from the ground with speed 20m/s. At time t = T seconds, the two balls are at the same vertical height, h metres, above the ground. The balls are modelled as particles moving freely under gravity.

I've got to the point where I have two equations:

Ball A -
h-50=2T+1/2(-9.8)t2

Ball B -
h=20T+1/2(-9.8)t2

What do I have to do now to solve h and T?
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#2
Any help will be appreciated!
0
4 years ago
#3
(Original post by ℓove)
At time t = 0, two balls A and B are projected vertically upwards. The ball A is projected vertically upwards with speed 2m/s from a point 50m above the horizontal ground. The ball B is projected vertically upwards from the ground with speed 20m/s. At time t = T seconds, the two balls are at the same vertical height, h metres, above the ground. The balls are modelled as particles moving freely under gravity.

I've got to the point where I have two equations:

Ball A -
h-50=2T+1/2(-9.8)t2

Ball B -
h=20T+1/2(-9.8)t2

What do I have to do now to solve h and T?
This is simply simultaneous equations, and so, assuming your equations are correct, we can subtract Ball A from Ball B, giving:

h-(h-50)=20T+1/2(-9.8)T2 -(2T+ 1/2(-9.8)T2)

50=18T

T=50/18

Then we can solve

h=20*T+1/2(-9.8)T2 plugging in our T value

1
#4
(Original post by mathematigeek)
This is simply simultaneous equations, and so, assuming your equations are correct, we can subtract Ball A from Ball B, giving:

h-(h-50)=20T+1/2(-9.8)T2 -(2T+ 1/2(-9.8)T2)

50=18T

T=50/18

Then we can solve

h=20*T+1/2(-9.8)T2 plugging in our T value

Thanks!!!!!
1
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