A 0.5 kg rock is projected from the edge of the top of a building with an initial velocity of 9.66 m/s at an angle of 41 degrees. The building is 10.1 m in height. At what horizontal distance, x, from the base of the building will the rock strike the

ground? Assume the ground is level and
that the side of the building is vertical. The acceleration of gravity is 9.8 m/s^2. Answer in units of m.

To find the horizontal distance at which the rock will strike the ground, we need to analyze the motion of the rock in both the horizontal and vertical directions.

First, let's consider the vertical motion. We can use the kinematic equation:

y = y0 + v0y * t - (1/2) * g * t^2

Where:
y = final vertical position (0, as the rock strikes the ground)
y0 = initial vertical position (10.1 m, the height of the building)
v0y = initial vertical velocity (v0 * sin(theta), where v0 is the initial velocity and theta is the launch angle)
g = acceleration due to gravity (9.8 m/s^2)
t = time (unknown)

In this case, we want to find the time at which the rock hits the ground, so we set y = 0 and rearrange the equation:

0 = 10.1 + (9.66 * sin(41)) * t - (1/2) * 9.8 * t^2

This is a quadratic equation that can be solved to find the value of t. Let's solve it:

0 = 10.1 + 6.47t - 4.9t^2

Rearranging, we get:

4.9t^2 - 6.47t - 10.1 = 0

Solving this quadratic equation, we find two possible values for t: t1 and t2.

Next, let's consider the horizontal motion. The horizontal velocity (v0x) remains constant throughout the motion, and it can be calculated using the initial velocity and the launch angle:

v0x = v0 * cos(theta)

In this case, v0x = 9.66 * cos(41).

The horizontal distance traveled (x) is equal to the horizontal velocity multiplied by the time:

x = v0x * t

Given the values for v0x and the two possible values for t, we can calculate two possible values for x: x1 and x2.

For the rock to hit the ground, it must be the case that both t and x are positive. Therefore, we need to select the values of t and x that meet these criteria.

By solving the quadratic equation and performing the calculations, we can find the horizontal distance from the base of the building at which the rock will strike the ground.

since you did not specify the direction very well (41 deg from horizontal or vertical? up or down from horizontal? Up or down from vertical?) I will have to be very general in the answer.

Vi = 9.66 sin or cos or -sin or -cos of 41

v = Vi - g t
at top v = 0
0 = Vi -9.81 t
so
t = Vi/9.81 at top

find h at top
h = 10.1 + Vi t - 4.9 t^2
solve for h, you know Vi and t

Now the rock drops from height h
0 = h + 0*Tfall - 4.9 Tfall^2
solve for Tfall, the time to fall from h

Now you know the rock has been in the air a time = t + Tfall

It had the same horizontal velocity the whole time

so
x = u ( t + Tfall)

where u = 9.66 cos or sin or -sin of 41