Find Speed of Stone at 10m: Solving for Speed at 10m from Ground

[erox]

Homework Statement

I have this question regarding how i would do the second part of this question as i already know how to do the first part ,but i can't seem to find an approach to the second."What is the speed of the stone at a height of 10m from the ground?"

A stone is thrown off the edge of a cliff and lands 37.5m from the base of the cliff.If the stone's initial speed was 11.5m/s,how high is the cliff?
What is the speed of the stone at a height of 10m from the ground?

Homework Equations

The Attempt at a Solution

tx=dx/vx
37.5/11.5m/s =3.26 secs

y=1/2gty^2
y=.5(9.8m/s^2)(3.26s)^2 = 52 m

 

[gash789]

Firstly, the question should have states "how" the stone was launched. I will assume it was horizatontally thrown with the given speed. If this is the case, then you must split the problem into two orthogonal directions, the obvious choice being x,y (where y is vertical).

Then you know the distance and velocity in the x direction and have correctly identified the time between it being launched and it striking the ground:

tx=dx/vx
37.5/11.5m/s =3.26 secs

you have then used
$y=\frac{1}{2} g t_{y}^{2}$
but for simplicity we can remember that tx=ty otherwise the situation would be very strange, and so
y=53m

What is the speed of the stone at a height of 10m from the ground?

Okay so now you need a function which will give you the velocity of the stone as a function of some other things, since you need not include any relevant equations, namely the equations of motion http://http://en.wikipedia.org/wiki/Equations_of_motion I think you will find all you looking for in equation 5

$v^{2}=u^{2}+2as$

 

[erox]

Firstly, the question should have states "how" the stone was launched. I will assume it was horizatontally thrown with the given speed. If this is the case, then you must split the problem into two orthogonal directions, the obvious choice being x,y (where y is vertical).

Then you know the distance and velocity in the x direction and have correctly identified the time between it being launched and it striking the ground:


you have then used
$y=\frac{1}{2} g t_{y}^{2}$
but for simplicity we can remember that tx=ty otherwise the situation would be very strange, and so
y=53m

What is the speed of the stone at a height of 10m from the ground?

Okay so now you need a function which will give you the velocity of the stone as a function of some other things, since you need not include any relevant equations, namely the equations of motion http://http://en.wikipedia.org/wiki/Equations_of_motion I think you will find all you looking for in equation 5

$v^{2}=u^{2}+2as$

Ok so using the equation you gave me i subtracted the height 52m -10m from the ground resulting as 42.1m.

v2^2=11.5m/s^2+2(9.8)(42.1m)

v2=30.94 m/s Would this be correct or would i have to a use the range equation for any reason R=x^2+y^2?

 

[gash789]

I am unsure what you mean by the range equation, but from it's form I would guess you mean
r^2=x^2+y^2 , that is Pythag' which would be the square of the physical distance the rock lands away.

You are asked for the "speed" which I am sure you are aware is a magnitude of the velocity (a vector). So you have calculated the y component of the velocity, and you know the x component (since gx=0) therefore you must calculate the magnitude of the velocity.

 

[asteriatic]

For the second part of the question, you can use the equation v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration due to gravity (9.8 m/s^2), and s is the displacement (in this case, 10m). Rearrange the equation to solve for v and plug in the known values:

v = √(u^2 + 2as)
v = √(11.5^2 + 2*9.8*10)
v = √(132.25 + 196)
v = √328.25
v = 18.12 m/s

Therefore, the speed of the stone at a height of 10m from the ground is 18.12 m/s.