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[parent] constant acceleration problems (Example)

Problem 1

At time $t = 0 \, sec$ we throw a stone from the ground level straight up with speed of $20 \, m/sec$ (ignore air drag, and assume $g = 10 \, m/sec^2$).

a) At what time (in sec) will this stone reach its highest point, and how high is it then above the ground?

b) We now throw a second stone straight up 2 sec after the first. How many meters above the ground is the first stone at that moment?

c) At what speed should we throw this second stone from the ground if it is to hit the first stone 1 second after the second stone is thrown?

Ans a)

Let us choose the positive y axis extending into the air. Then the force due to gravity is in the negative y direction

$\displaystyle F = -mg $

applying Newton's 2nd law

$\displaystyle F = m\ddot{y} = -mg $

Integrate this to get equation for velocity of the stone

$\displaystyle \dot{y}(t) = -g t + C_1 $

At $t = 0, \, \dot{y} = 20 \, m/s$, therefore

$\displaystyle \dot{y}(t) = -g t + 20$ (1)

Integrate again to get equation for position of the stone

$\displaystyle y(t) = -\frac{1}{2} g t^2 + 20t + C_2 $

Once, again plug in the initial condition that at $t = 0, \, y = 0$ to get $C_2$

$\displaystyle 0 = C_2$

therefore

$\displaystyle y(t) = -\frac{1}{2}g t^2 + 20t$ (2)

Next we know that at the stone's highest point $h$ its velocity will be 0. So from equation (1) we can then find the time $t_h$ at this point

$\displaystyle 0 = -gt_h + 20 $

$\displaystyle t_h = 20 / g = 20 / 10 = 2 \, sec $

Now we can plug this result into equation (2) to get the height

$\displaystyle h = -\frac{1}{2}g t_h^2 + 20t_h = -\frac{1}{2}(10)(2^2) + 20(2)$

Therefore

$\displaystyle h = 20 \, [m] $

Ans b)

This is straight forward since equation 2 gives us the position of the first stone at some given time. So at $t = 2$

$\displaystyle y(2) = -\frac{1}{2}g(2)^2 + 20(2) = 40 - 20 = 20 \, [m] $

which, makes sense from question a.

Ans c)

The stones will hit when first stone is 3 seconds into its flight. Its height is then determined from equation 2

$\displaystyle y = -\frac{1}{2}(10)(3)^2 + 20 (3) = -45 + 60 = 15 \, [m] $

If we now look back to see how we got the constant in equation (1), we see that $C_1$ is equal to the initial velocity $v_0$

$\displaystyle \dot{y} = -gt + v_0 $

Integrate to get position

$\displaystyle y = -\frac{1}{2} gt^2 + v_0 t $

We want to find $v_0$ at 1 second into the second stone's flight which we know occurs at $15 \, [m]$. So set $y=15$ and solve for $v_0$ at $t=1$

$\displaystyle 15 = -\frac{1}{2} (10)(1)^2 + v_0(1)$

$\displaystyle 15 = -5 + v_0$

So we see that the initial velocity needed for second stone is

$\displaystyle v_0 = 20 \, [m/s] $

There is another way of finding the speed without making any calculations. At $t=3,$ the first stone is at the same height as it was at $t=1 \, sec.$ Since the stones have to collide at this height exactly 1 sec after the second stone is thrown, the second stone should also begin with a speed of 20 m/sec.

References

This is a derivative work from [1] a Creative Commons Attribution-Noncommercial-Share Alike 3.0 work.

[1] MIT OpenCourseWare, 8.01 Physics I: classical mechanics, Fall 1999.



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Cross-references: classical mechanics, work, position, velocity, speed

This is version 3 of constant acceleration problems, born on 2009-02-11, modified 2009-02-11.
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Classification:
Physics Classification: 45.50.Dd (General motion)
 45.50.Pk (Celestial mechanics )
 45.50.-j (Dynamics and kinematics of a particle and a system of particles)
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