Finding Vertical Distance of a Horizontally Thrown Ball at 45.08 m/s

[Warwick]

I just started doing some problems and this basic pm problem is giving me a lot of trouble.

If a ball was throwen horizontally with 45.08 m/s, how far would the ball fall vertically by the time it reached home plate, 18.28m away?

The initial velocity is 45.08 m/s. Now that is the x and y velocity combined, correct? How do I go about finding the x and y velocity? If I don't need to do it for this particular problem could you explain how to do it anyway and how to solve the problem. I tried using arc-cos to find the initial angle then find the velocities from there but I don't think that's right.

Thanks for reading.

 

[Nylex]

If it's thrown horizontally, the vertical component of its initial velocity is 0. Since there aren't any forces acting on the ball horizontally (ignoring air resistance), it travels at a constant velocity. You can then work out the time taken to travel 18.28 m (t = x/v). What you want to do is find out the distance the ball falls in this time. Use the equation s = ut + (1/2)at^2 (using a = g, u = 0 as explained above and t as the time you've calculated).

 

[wais]

To find the vertical distance of a horizontally thrown ball, we can use the equation d = 1/2 * g * t^2, where d is the vertical distance, g is the acceleration due to gravity (9.8 m/s^2), and t is the time the ball is in the air.

In this case, we know the initial velocity (45.08 m/s) and the horizontal distance (18.28 m). To find the time, we can use the equation d = v*t, where d is the horizontal distance and v is the initial velocity. Rearranging this equation, we get t = d/v.

Plugging in the values, we get t = 18.28 m / 45.08 m/s = 0.406 seconds.

Now, we can plug this value of time into the first equation to find the vertical distance: d = 1/2 * 9.8 m/s^2 * (0.406 s)^2 = 0.8 m.

So, the ball would fall 0.8 meters vertically by the time it reaches home plate.

To find the x and y velocities, we can use trigonometry. The initial velocity is the hypotenuse of a right triangle, with the x and y velocities as the legs. We can use the formula v = v0*cos(theta) for the x velocity and v = v0*sin(theta) for the y velocity, where v0 is the initial velocity and theta is the angle at which the ball is thrown.

To find theta, we can use the inverse tangent function, since we know the opposite (y velocity) and adjacent (x velocity) sides of the triangle. So, theta = tan^-1 (y velocity / x velocity).

In this case, since the ball is thrown horizontally, the y velocity is 0. So, theta = tan^-1 (0 / x velocity) = 0.

This means that the x velocity is equal to the initial velocity (45.08 m/s) and the y velocity is 0.

I hope this helps with your understanding of the problem. Keep practicing and you'll become more comfortable with these types of calculations!