Vertical motion under gravity - SUVAT equations.

[rollcast]

Homework Statement

Gravity is assumed to be 10ms^-2 throughout.

A ball is thrown vertically upward with a speed of 14ms^-1. Two seconds later, a ball is dropped from the same point find where the two balls meet.

(They will meet when they are the same displacement (s) from the starting point. If the time taken for ball 1 is T, then the time taken for the second ball is T - 2 seconds.)

Homework Equations

v=u+at
v²=u²+2as
s=ut+(1/2)at²
s=(1/2)(u+v)t

The Attempt at a Solution

The answer is 8.\dot{8} and the teacher showed another method for solving it but I felt I was getting somewhere with my method but I couldn't figure out what to do next.

At t=0 the variables for ball 1 are:

s=0m
u=14ms^-1
a=-10ms^-2 as the force of gravity is opposing motion.

So then I can work out the velocity and displacement of ball one at t=2.

v=u+at
v=14-10*2
v=-6ms^-1 so its moving downwards.

s=(1/2)(u+v)*t
s=(1/2)(14-6)*2
s=(1/2)(8)*2
s=8m above the starting position.

This is really as far as I got any meaningful answers as I couldn't figure out what to do next so I just copied down the example method from the black board.

Thanks
Al

PS. I can do a couple of diagrams if anyone needs them to visualise the problem.

 

[PeterO]

Homework Statement

Gravity is assumed to be 10ms^-2 throughout.

A ball is thrown vertically upward with a speed of 14ms^-1. Two seconds later, a ball is dropped from the same point find where the two balls meet.

(They will meet when they are the same displacement (s) from the starting point. If the time taken for ball 1 is T, then the time taken for the second ball is T - 2 seconds.)

Homework Equations

v=u+at
v²=u²+2as
s=ut+(1/2)at²
s=(1/2)(u+v)t

The Attempt at a Solution

The answer is 8.\dot{8} and the teacher showed another method for solving it but I felt I was getting somewhere with my method but I couldn't figure out what to do next.

At t=0 the variables for ball 1 are:

s=0m
u=14ms^-1
a=-10ms^-2 as the force of gravity is opposing motion.

So then I can work out the velocity and displacement of ball one at t=2.

v=u+at
v=14-10*2
v=-6ms^-1 so its moving downwards.

s=(1/2)(u+v)*t
s=(1/2)(14-6)*2
s=(1/2)(8)*2
s=8m above the starting position.

This is really as far as I got any meaningful answers as I couldn't figure out what to do next so I just copied down the example method from the black board.

Thanks
Al

PS. I can do a couple of diagrams if anyone needs them to visualise the problem.

If you want to just talk (think out loud may be a better description) through the problem [using g = 10 makes this possible] we have.

The stone thrown up at 14m/s will dake 1.4 seconds to stop. In a further 1.4 seconds, it will be back where it started - this time traveling DOWN at 14 m/s - so a total of 2.8 seconds.

The dropped stone thus has a 0.8 second head start in the downward race.
By then it will have reached 8 m/s, and traveled 0.8*(0+8)/2 = 3.2m

From the 2.8 second mark, we have two stones, both accelerating under the influence of gravity. The first stone is traveling at 14 m/s, the dropped stone traveling at 8 m/s. That means the first stone is gaining at 6 m/s - and has to make up 3.2 m.

That will take 0.5333333 seconds.

That mans 1.33333333 seconds after the stone was dropped.

How far does a ball fall in 4/3 seconds - because that is where they meet.

in 4/3 seconds, it will reach 40/3 m/s, and so will have averages 20/3 m/s
It has averaged that for 4/3 seconds, covering a distance of 80/9 m.

If only the original standard metre had been constructed just a little bit shorter; the real value of g would be 10 and all calculations for vertical and projectile motion would be this easy ! ( we wouldn't have to use 9.81, and resort to our calculator)

 

[PeterO]

Homework Statement

Gravity is assumed to be 10ms^-2 throughout.

A ball is thrown vertically upward with a speed of 14ms^-1. Two seconds later, a ball is dropped from the same point find where the two balls meet.

(They will meet when they are the same displacement (s) from the starting point. If the time taken for ball 1 is T, then the time taken for the second ball is T - 2 seconds.)

Homework Equations

v=u+at
v²=u²+2as
s=ut+(1/2)at²
s=(1/2)(u+v)t

The Attempt at a Solution

The answer is 8.\dot{8} and the teacher showed another method for solving it but I felt I was getting somewhere with my method but I couldn't figure out what to do next.

At t=0 the variables for ball 1 are:

s=0m
u=14ms^-1
a=-10ms^-2 as the force of gravity is opposing motion.

So then I can work out the velocity and displacement of ball one at t=2.

v=u+at
v=14-10*2
v=-6ms^-1 so its moving downwards.

s=(1/2)(u+v)*t
s=(1/2)(14-6)*2
s=(1/2)(8)*2
s=8m above the starting position.

This is really as far as I got any meaningful answers as I couldn't figure out what to do next so I just copied down the example method from the black board.

Thanks
Al

PS. I can do a couple of diagrams if anyone needs them to visualise the problem.

Using the discussion in my previous post - and the two figures you calculate above:

If we take the "race" to begin at the 2 second mark then.
The first ball is starting 8m above the second, but is traveling at 6m/s faster.

Since they have the same acceleration (g) the first ball will always be traveling at 6 m/s faster.

at 6 m/s, it takes 8/6 (or 4/3) seconds to catch up.

How far does the dropped ball travel in 4/3 seconds?