My problem question:

"A ball is thrown up onto a roof, landing 4.0 seconds later at a height of 20.0 m above where it was thrown. The ball's velocity vector upon landing on the roof top was angled 60 degrees to the horizontal flat roof top. Find the horizontal distance that the ball traveled."

Which equation should I use to find that distance?

140

To find the horizontal distance that the ball traveled, you can use the equation for horizontal displacement:

Horizontal displacement = Velocity × Time

Given that the ball's velocity vector upon landing on the roof was angled 60 degrees to the horizontal, the horizontal component of the velocity can be found using the equation:

Horizontal velocity = Velocity × cos(θ)

Where θ is the angle between the velocity vector and the horizontal.

Therefore, the equation you should use to find the horizontal distance traveled by the ball is:

Horizontal displacement = Horizontal velocity × Time

Substituting the equation for horizontal velocity, the equation becomes:

Horizontal displacement = (Velocity × cos(θ)) × Time

To find the horizontal distance that the ball traveled, we can use the equation of motion for horizontal motion. This equation relates the horizontal distance traveled (d), the initial horizontal velocity (v₀x), and the time of flight (t). The equation is:

d = v₀x * t

In this case, we need to determine the initial horizontal velocity, v₀x. Since the ball's velocity vector upon landing on the roof top was angled 60 degrees to the horizontal flat roof, we can use trigonometry to find the horizontal component of the velocity.

We know that the angle of 60 degrees is opposite to the vertical component of the velocity, which we'll call v₀y, and the hypotenuse of the triangle formed by the velocity vector is the magnitude of the initial velocity, which we'll call v₀.

Using trigonometry, we can determine:

v₀x = v₀ * cos(theta)
where theta is the angle made with the horizontal flat roof, which we know is 60 degrees in this case.

Now, we have all the values needed to substitute into the equation for horizontal distance:

d = (v₀ * cos(theta)) * t

Plug in the given values of t = 4.0 seconds, theta = 60 degrees, and the magnitude of the initial velocity v₀ (which is not given in the problem) to solve for the horizontal distance traveled by the ball.

You need more than one equation to solve this problem. From the time of flight that it takes to land on the roof, you can calculate the initial vertical component of the launch velocity, Vyo.

20.0 m = Vyo *4.0 - (g/2)*4.0^2
20 + 78.4 = 98.4 m = 4 Vyo
Vyo = 24.6 m/s
The vertical velocity component when it hits the roof is
Vy = Vyo - gt = 24.6 - 39.2 = -14.6 m/s
(It is negative because it is travelling down).

From the trajectory angle when it hits the roof,
Vy/Vx = -tan60 = -1.732
Vx = 8.43 m/s
The Vx velocity component remains constant, so it also equals Vxo at launch.

The distance traveled horizontally is Vx*4.0 s = ___