Physics Problem Help: Calculating Horizontal Velocity for Pebble Toss

[kimmeh06]

1.Homework Statement

1. You want to toss the pebbles up to a friend's window such that the pebbles hit the window with only a horizontal component of velocity. The window is 8.00m above you and there is a 6.00m wide flowerbed between you and the wall of the house. A) How fast do you throw the pebble? B) How fast are the pebbles going when they hit your friend's window?2. Relative Equations
2. X=Xo+Vxo*t Y=Yo+Vyot+.5at^23. Attempt at the solution.
3. I know that we have to solve for initial velocity and then final velocity with only the horizontal (x) component. What I did was use trig to get tan(theta)=8/6 to get theta=53degrees. From there I plugged the variables into the equations to get
8=Vosin(theta)*t-.5*g*t^2 and 6=Vcos(theta)*t From here I'm not really sure what to do or even if I set up the equation right. Any help would be soooo appreciated! Thank you.

 

[sir_manning]

Image you are standing right below the window, and you want to throw the rock directly upward so that it *just* reaches the window at the top of its flight. The velocity at which you have to throw the pebble for this to happen will be the y-component of velocity in your situation. Since you don't have t, try using V²=Vo²+2*a*d to calculate this.

Then once you know what the y-component of velocity is, you should be able to find the x-component easily.

 

[kerimek]

Dear student,

Thank you for reaching out for help with this physics problem. It seems like you have made a good start by using trigonometry to find the angle at which you need to throw the pebble. However, there are a few things to keep in mind when solving this type of problem.

Firstly, it is important to remember that the initial velocity (Vx) and final velocity (Vx') in the horizontal direction will be the same, as there is no acceleration acting on the pebble in this direction. This means that we can use the same value for Vx in both equations.

Secondly, we need to consider the vertical motion of the pebble as well. The pebble will be launched with an initial vertical velocity (Vy) and will follow a parabolic trajectory due to the acceleration of gravity (g). This means that we need to use the equations for both horizontal and vertical motion to solve for the initial and final velocities.

With these considerations in mind, we can set up the equations as follows:

Equation 1: X = Xo + Vxo*t
Equation 2: Y = Yo + Vyot + 0.5at^2

Where:
X = final horizontal position (6.00m)
Xo = initial horizontal position (0m)
Vxo = initial horizontal velocity (unknown)
t = time taken for the pebble to travel horizontally (unknown)
Y = final vertical position (8.00m)
Yo = initial vertical position (0m)
Vyo = initial vertical velocity (unknown)
a = acceleration due to gravity (-9.8m/s^2)

Now, using the trigonometric relationship you have found (tan(theta) = 8/6), we can substitute in the values for Vxo and Vyo as follows:

Vxo = Vx*cos(theta)
Vyo = Vx*sin(theta)

Substituting these into our equations, we get:

Equation 1: 6 = Vx*cos(theta)*t
Equation 2: 8 = Vx*sin(theta)*t - 0.5*9.8*t^2

Now, we have two equations with two unknowns (Vx and t). We can solve these simultaneously to find the values for Vx and t.

To do this, we can rearrange equation 1 to solve for t:

t = 6/Vx*cos(theta)

We can then substitute this