# Basic mechanics problem

• Jun 11th 2009, 04:09 PM
AshleyT
Basic mechanics problem
Okey, my exam is in four days and im officially freaking out! I seem to have forgotten everything having been revising for other exams...urg...can't even do basic mechanics questions!

So any help would be greatly, greatly appreciated!

Stone is thrown Horizontally with speed 15m/s from top of a cliff, 30 meters high. Estimate the distance of the stone from the thrower when it is level with the foot of the cliff and the time it takes for the stone to fall.

Now, wth? With the given information all we have is U = 15m/s...how is it possible to find S?

Thanks!
• Jun 11th 2009, 04:30 PM
Chris L T521
Quote:

Originally Posted by AshleyT
Okey, my exam is in four days and im officially freaking out! I seem to have forgotten everything having been revising for other exams...urg...can't even do basic mechanics questions!

So any help would be greatly, greatly appreciated!

Stone is thrown Horizontally with speed 15m/s from top of a cliff, 30 meters high. Estimate the distance of the stone from the thrower when it is level with the foot of the cliff and the time it takes for the stone to fall.

Now, wth? With the given information all we have is U = 15m/s...how is it possible to find S?

Thanks!

At $t=0$, the stone is thrown horizontally off the cliff with the following velocities: $v_{0x}=15\,m/s,\,v_{0y}=0\,m/s$. Let $g=9.8\,m/s^2$, $y_0=30\,m$, $y_f=0\,m$, $x_0=0\,m$ and $x_f=?$

From Kinematics, we know that $\Delta y=v_{0y}t-\tfrac{1}{2}gt^2\implies -30m=\left(-4.9\,m/s^2\right)t^2\implies t^2=6.12\,s^2\implies t\approx 2.47\,s$

We also know that $\Delta x= v_{0x}t=15\, m/s\cdot 2.47\,s=37.11\,m$

Thus, the stone travels $37.11\,m$ in the horizontal direction, doing so in $2.47\,s$

Does this make sense?
• Jun 11th 2009, 04:42 PM
AshleyT
Quote:

Originally Posted by Chris L T521
At $t=0$, the stone is thrown horizontally off the cliff with the following velocities: $v_{0x}=15\,m/s,\,v_{0y}=0\,m/s$. Let $g=9.8\,m/s^2$, $y_0=30\,m$, $y_f=0\,m$, $x_0=0\,m$ and $x_f=?$

From Kinematics, we know that $\Delta y=v_{0y}t-\tfrac{1}{2}gt^2\implies -30m=\left(-4.9\,m/s^2\right)t^2\implies t^2=6.12\,s^2\implies t\approx 2.47\,s$

We also know that $\Delta x= v_{0x}t=15\, m/s\cdot 2.47\,s=37.11\,m$

Thus, the stone travels $37.11\,m$ in the horizontal direction, doing so in $2.47\,s$
Does this make sense?

Ah sorry, we've not done projectiles to that extent. The only knowledge we have is SUVAT equations and the basic projectile motion.

But thankyou very much for the reply!
• Jun 12th 2009, 03:55 AM
mr fantastic
Quote:

Originally Posted by AshleyT
Ah sorry, we've not done projectiles to that extent. The only knowledge we have is SUVAT equations and the basic projectile motion.

But thankyou very much for the reply!

Motion in the vertical direction (take downwards as the +ve direction):

a = 9.8 m/s^2
s = 30 m
u = 0 m/s
t = ?

Find t using $s = ut + \frac{1}{2} a t^2$.

Motion in the horizontal direction:

a = 0 m/s^2
u = 15 m/s
t = value calculated above.
s = ?

Find s using $s = ut + \frac{1}{2} a t^2$. This is the distance from the base of the cliff, which is not the same as the distance of the stone from the thrower.

If you want the distance of the stone from the thrower, then you need to use Pythagoras' Theorem: $d^2 = 30^2 + s^2$ where d is the distance and s is the value calculated above.
• Jun 14th 2009, 04:08 AM
AshleyT
Quote:

Originally Posted by mr fantastic
Motion in the vertical direction (take downwards as the +ve direction):

a = 9.8 m/s^2
s = 30 m
u = 0 m/s
t = ?

Find t using $s = ut + \frac{1}{2} a t^2$.

Motion in the horizontal direction:

a = 0 m/s^2
u = 15 m/s
t = value calculated above.
s = ?

Find s using $s = ut + \frac{1}{2} a t^2$. This is the distance from the base of the cliff, which is not the same as the distance of the stone from the thrower.

If you want the distance of the stone from the thrower, then you need to use Pythagoras' Theorem: $d^2 = 30^2 + s^2$ where d is the distance and s is the value calculated above.

If i could thankyou even more so...i would =)
Thankyou very much.
• Jun 14th 2009, 04:13 AM
mr fantastic
Quote:

Originally Posted by AshleyT
If i could thankyou even more so...i would =)
Thankyou very much.

You're welcome. Good luck with the exam.