Solve Ball Vector Problem: Distance, Vertical & Horizontal Components

[mossfan563]

Homework Statement

You throw a ball toward a wall at speed 37.0 m/s and at angle θ0 = 43.0° above the horizontal (Fig. 4-35). The wall is distance d = 20.0 m from the release point of the ball. (a) How far above the release point does the ball hit the wall? What are the (b) horizontal and (c) vertical components of its velocity as it hits the wall?

Homework Equations

V_x = V_0 cos(theta)
V_y = V_0 sin(theta) - gt

The Attempt at a Solution

I was able to find b by using V_x = V_0 cos(theta) with theta being 43 degrees and V_0= 37.
When I tried to find part a, I thought it was a simple tan association.
tan (43) = x/20 -> x = 20*tan(43)
I got x = 18.65... and it was wrong.
I tried getting part c using V_y = V_0 sin(theta) - gt with:
V_0 = 37, theta = 43 and g = 9.8
Then I got stuck.
Am I doing it right? How do you approach a and c?

 

[Redbelly98]

When I tried to find part a, I thought it was a simple tan association.
tan (43) = x/20 -> x = 20*tan(43)
I got x = 18.65... and it was wrong.

That would work if the ball traveled in a straight line, at constant velocity. But that doesn't happen because of gravity.

I tried getting part c using V_y = V_0 sin(theta) - gt with:
V_0 = 37, theta = 43 and g = 9.8
Then I got stuck.
Am I doing it right? How do you approach a and c?

That's good, but you need to find t to finish the question.

Can you use the horizontal distance and velocity to find t?

 

[baba89]

To solve this problem, we can use the equations for horizontal and vertical components of velocity, V_x = V_0 cos(theta) and V_y = V_0 sin(theta) - gt, where V_0 is the initial velocity, theta is the angle of release, and g is the acceleration due to gravity.

For part a, we need to find the horizontal distance traveled by the ball before hitting the wall, which is given by the equation x = V_x * t, where t is the time it takes for the ball to reach the wall. To find t, we can use the vertical component of velocity equation, V_y = V_0 sin(theta) - gt, and solve for t. This gives us t = (V_0 sin(theta))/g. Now, substituting this value of t in the equation x = V_x * t, we get x = V_0 cos(theta) * (V_0 sin(theta))/g. Plugging in the values given in the problem, we get x = 20.22 m. Therefore, the ball hits the wall at a horizontal distance of 20.22 m from the release point.

For part b, we can use the equation V_x = V_0 cos(theta) to find the horizontal component of velocity. Plugging in the values given in the problem, we get V_x = 31.28 m/s.

For part c, we can use the equation V_y = V_0 sin(theta) - gt to find the vertical component of velocity. Plugging in the values given in the problem, we get V_y = 14.27 m/s.

Therefore, the ball hits the wall at a horizontal distance of 20.22 m from the release point, with a horizontal velocity of 31.28 m/s and a vertical velocity of 14.27 m/s.