A baseball thrown at an angle of 55.0 degrees above the horizontal strikes a building 16.0 m away at a point 8.00 m above the point from which it is thrown. Ignore air resistance.
Find the magnitude of the initial velocity of the baseball (the velocity with which the baseball is thrown).
got it.... 16.0 m/s
Find the magnitude of the velocity of the baseball just before it strikes the building.
I have tried so many things it isn't even funny. HELP!
Find the direction of the velocity of the baseball just before it strikes the building.
Yeah right...like I know....
Did you set up the position and velocity vectors for the problem? That would help.
The position vector is
(x,y)=(cos(55)*v*t, (-1/2)g*t^2+sin(55)*v*t+yo)
Using the x component you see that
(1) 16m=cos(55)*v*t
For the y component we're told y_f=yo+8=(-1/2)g*t^2+sin(55)*v*t+yo
This means
(-1/2)g*t^2+sin(55)*v*t - 8 = 0
From (1) we can use 16m/cos(55) = v*t
Since you know g you can you can solve for t.
Once you know t you can determine v from (1) (Verify that I've given the correct method to do this. I see you have an answer already.)
The velocity vector for the ball is
(2) (v_x,v_y)=)=(cos(55)*v, -1g*t+sin(55)*v)
The magnitude of the velocity is the speed and it's given by
(3) Speed=sqrt((v_x)^2 + (v_y)^2)
Since you should now have the time and the inital velocity, calculate v_x and v_y and use (3) to find the speed.
Use the time you found and (2) to give the velocity vector for the ball.
If you use a diagram be sure to set up the coordinate system so you're throwing from left to right and yo is the height at the origin.
Repost if you have questions.
How did you find the initial velocity without the time?
Well, aren't you getting all technical with position vectors and velocity vectors! It's like you're speaking a whole different language! But don't worry, I'm here to help, even if I don't understand half of what you just said.
Let's break it down in a way that your average clown would understand. To find the magnitude of the initial velocity, you can use the equation 16m = cos(55)*v*t. But if you're like me and numbers make your head spin, just remember that the answer is 16 m/s. Easy peasy!
Now, let's move on to the magnitude of the velocity just before the ball strikes the building. Forget about all those position vectors and equations for now. The magnitude of the velocity is simply the speed, which can be found using the equation Speed = sqrt((v_x)^2 + (v_y)^2). In this case, the vertical component of the velocity is -1g*t+sin(55)*v, and the horizontal component is cos(55)*v. Plug those into the equation and you'll get your answer.
Lastly, the direction of the velocity just before the ball strikes the building. Well, who am I to tell you the direction? You're the one who set up the coordinate system with x and y and whatnot. But my best guess would be that it's probably headed towards the building. Just a hunch!
I hope that helps, even if my clown brain can't fully comprehend all the technical stuff. Remember, sometimes a little humor goes a long way in solving tricky problems!
To find the magnitude of the velocity of the baseball just before it strikes the building, you can use the equation (3) mentioned above:
Speed = sqrt((v_x)^2 + (v_y)^2)
We already have the values for v_x and v_y from equation (2):
v_x = cos(55) * v
v_y = -gt + sin(55) * v
To find the value of t, we can use the equation:
(-1/2)g*t^2 + sin(55)*v*t - 8 = 0
Solving this equation will give us the value of t. Once we have t, we can substitute it into equations (2) to find v_x and v_y.
Finally, substitute the values of v_x and v_y into equation (3) to calculate the speed, which is the magnitude of the velocity just before the baseball strikes the building.
As for the direction of the velocity, you can find it by using the inverse tangent function:
Direction = arctan(v_y/v_x)
This will give you the angle of the velocity vector with respect to the positive x-axis.
To find the magnitude of the initial velocity of the baseball (the velocity with which the baseball is thrown), you can use the position vector equations provided. The x component of the position vector equation (x = cos(55) * v * t) gives us the distance the baseball travels horizontally (16m). Using this equation, we can rearrange it to solve for t:
t = 16m / (cos(55) * v)
Next, we can substitute this value of t into the y component of the position vector equation (-1/2 * g * t^2 + sin(55) * v * t - 8 = 0) to solve for v:
-1/2 * g * (16m / (cos(55) * v))^2 + sin(55) * v * (16m / (cos(55) * v)) - 8 = 0
Solving this equation will give you the value of v, which is the magnitude of the initial velocity of the baseball.
To find the magnitude of the velocity of the baseball just before it strikes the building, we can use the velocity vector equation (v_x = cos(55) * v) and (v_y = -g * t + sin(55) * v) with the value of t we just calculated. Plug in the value of t into the velocity vector equation and use the magnitude formula (sqrt((v_x)^2 + (v_y)^2)) to find the magnitude of the velocity.
To find the direction of the velocity of the baseball just before it strikes the building, you can use the value of the x and y components of the velocity vector (v_x and v_y). The direction can be determined using the tangent function:
direction = arctan(v_y/v_x)
This will give you the angle at which the baseball is moving just before it strikes the building.