Projectiles...

[fairylight]

I have a projectiles problem, ive been puzzling for hours and just cant seem to work it out! Its probably so simple but if someone could point me in the right direction please help!

 

A stone is hurled into the air on level ground with a horizontal component of velocity 30m/s and a vertical component of 20m/s. How long before the stone falls to the ground?

 

Do i have to find the resultant velocity, then the height before i can work it our or do i just use acceleration due to gravity? Im so confused!!! :confused:

[timo]

Do you think the horizontal velocity matters for the question how long the stone will be in the air?

[5614]

And remember that you know that acceleration will be gravity, so a = -9.8

 

It's negative assuming that upwards is positive and downwards is negative.

[fairylight]

so do i just do change in velocity / acceleration to get time?! like 20/9.81. surely theres more to it than that????!!!! lol im awful at physics!

[jordan]

Nope. You want to use these facts:

 

Initial Velocity=20m/sec

Final Velocity=0m/sec

Acceleration=-9.81m/sec²

Time:?

 

Find a way to relate these four facts. But note that I used a Final Velocity of 0. The rock wont hit the ground with a velocity of 0 so this is only solving part of the solution. You'll have to take another step after you get your answer from above to get the final answer.

[fairylight]

so wud i find the time it takes for the stone to reach its peak...using the equations of motion then just double for time to reach the ground? is it v=u+at ??????

[Severian]

How sophisticated is your maths? The most complete way to do this equation is the rigourous way.

 

Define [math]\vec{i}[/math] and [math]\vec{j}[/math] as unit vectors in the horizontal and vertical directions respectively. Then the initial velocity is [math]\vec{v}_0 = (30 \vec{i}+20\vec{j}) {\rm ms}^{-1}[/math], while the stone's acceleration is [math]\vec{a} = -g\vec{j} = - 9.81 \vec{j} {\rm ms}^{-2}[/math].

 

Now solve [math]\vec{a} = \frac{d\vec{v}}{dt}[/math] to find the velocity [math]\vec{v}[/math] as a funtion of time (using your initial condition on [math]\vec{v}_0[/math]).

 

Now write the position of the stone as [math]\vec{r}[/math] with an initial condition [math]\vec{r}_0 = 0[/math] and solve [math]\vec{v}(t)=\frac{d\vec{r}}{dt}[/math] for [math]\vec{r}[/math] as a function of time (using your initial condition on [math]\vec{r}_0[/math]).

 

Finally, your stone hits the ground when [math]\vec{r} \cdot \vec{j}=0[/math]. Solve this for [math]t[/math] (neglecting the trivial [math]t=0[/math] solution, and you have your answer.

[jordan]

so wud i find the time it takes for the stone to reach its peak...using the equations of motion then just double for time to reach the ground? is it v=u+at ??????

If you didn't understand what Severian just said, then yes, that's the best method for you to use.