Mechanics 1 Question on Constant Acceleration

A ball A is thrown vertically upwards at 25metres per second from a point P. Three seconds later a second ball B is also thrown vertically upwards from the point P at 25 metres per second. Taking the acceleration due to gravity to be 10 m/s^2, calculate the time for which ball A has been in motion when the balls meet.

I understand you need to use S=ut + 0.5 a t^2 and find the t value for when the distance is the same, but I'm not sure how to write the displacement equation for the second ball (because it is released after 3 seconds, I.e. Do you need to write the equation of a graph that consists of two straight lines, a flat line which is stationary for 3 seconds, which then rises.

Reply 1

s_ahmed2

13

Original post by Cyan Ink

A ball A is thrown vertically upwards at 25metres per second from a point P. Three seconds later a second ball B is also thrown vertically upwards from the point P at 25 metres per second. Taking the acceleration due to gravity to be 10 m/s^2, calculate the time for which ball A has been in motion when the balls meet.

I understand you need to use S=ut + 0.5 a t^2 and find the t value for when the distance is the same, but I'm not sure how to write the displacement equation for the second ball (because it is released after 3 seconds, I.e. Do you need to write the equation of a graph that consists of two straight lines, a flat line which is stationary for 3 seconds, which then rises.

Use two equations, for the second one(b) t=t+3

Posted from TSR Mobile

Reply 2

tiny hobbit

15

Original post by Cyan Ink

A ball A is thrown vertically upwards at 25metres per second from a point P. Three seconds later a second ball B is also thrown vertically upwards from the point P at 25 metres per second. Taking the acceleration due to gravity to be 10 m/s^2, calculate the time for which ball A has been in motion when the balls meet.

I understand you need to use S=ut + 0.5 a t^2 and find the t value for when the distance is the same, but I'm not sure how to write the displacement equation for the second ball (because it is released after 3 seconds, I.e. Do you need to write the equation of a graph that consists of two straight lines, a flat line which is stationary for 3 seconds, which then rises.

Original post by s_ahmed2

Use two equations, for the second one(b) t=t+3

Posted from TSR Mobile

The second ball is travelling for a smaller time, so if you take the time for the first ball as t, the second ball will take t - 3.

Reply 3

cliverlong

8

An alternate way to think about this is to draw a vt graph, where if the original velocity is on e positive, vertical axis, both lines will have negative gradient of -9.8. Then determine the equations for both lines. Devise an equation for the displacement by considering the area "under" the graphs and let the area when graph drops below the t axis become negative. Then when the displacement equations are equal this will give you the time of intersection. The equations produced will be the same as in the previous method, just a different way of getting there.

Mechanics 1 Question on Constant Acceleration

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