Back
to top
Help with M1 question
Watch
Announcements
Page 1 of 1
Go to first unread
Skip to page:
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?
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?
0
reply
Report
#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?
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?
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
Please note, I havent checked that your equations are correct.
1
reply
(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
Please note, I havent checked that your equations are correct.
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
Please note, I havent checked that your equations are correct.
1
reply
X
Page 1 of 1
Go to first unread
Skip to page:
Quick Reply
Today on TSR
-
Ofqual consultationTeachers to use 'mini-exams' and other work to decide grades
- Is it too late to do well in my A-levels?
- How are you feeling today?
- 2021 wellbeing blog launch
- Student tips on lockdown mental health