You attempt to throw the rock across the chasm with a velocity of 35 m/s at and angle of 60 degrees above horizontal. Calculate and label the forward velocity (components) of the rock as it leaves your hand. How long will it take to reach the top of the arc (apogee. How far away is the rock (from your hand) when is level with your hand the second time.

Try and draw a free body diagram to help you understand this. The 35 m/s velocity is along the hypotenuse of a right triangle, and the components you want are in the x and y direction. To find the y component, you would use sinA=y/35, and for the x component, you would use cosA=x/35.

To reach the top of the arch, use the motion equations. You now have the velocity in the y direction, and you know acceleration due to gravity is 9.8 m/s2. Final velocity in the y direction at the top of the arch is 0, so use the equation vf2=vi2-2(9.8)(y) and solve for y.
The rock will be equal with your hand when the velocity in the y direction is equal to the initial y velocity, but in the opposite direction, so y=0. Use the equation y=vt-(1/2)(9.8)t2 to find the time it takes, then plug that time into the equation for movement in the x direction, which is x=v(in the x direction)t. Remember velocity in the x direction does not change in the absence of air resistance.

To calculate the forward velocity components of the rock as it leaves your hand, we need to separate the velocity into its horizontal and vertical components.

The horizontal component, labeled Vx, can be obtained by multiplying the initial velocity (35 m/s) by the cosine of the launch angle (60 degrees).

Vx = V₀ * cos(θ)
Vx = 35 m/s * cos(60°)
Vx ≈ 17.5 m/s

The vertical component, labeled Vy, can be obtained by multiplying the initial velocity (35 m/s) by the sine of the launch angle (60 degrees).

Vy = V₀ * sin(θ)
Vy = 35 m/s * sin(60°)
Vy ≈ 30.2 m/s

So, the forward velocity components of the rock as it leaves your hand are approximately Vx = 17.5 m/s and Vy = 30.2 m/s.

Now, let's calculate how long it will take for the rock to reach the top of the arc (apogee).

The time taken to reach the top of the arc can be determined using the vertical component of the velocity (Vy) and the acceleration due to gravity (g ≈ 9.8 m/s²). At the top of the arc, the vertical component of the velocity becomes zero.

Vy = V₀ * sin(θ) - g * t
0 = 35 m/s * sin(60°) - 9.8 m/s² * t

We can rearrange the equation to solve for time (t).

t = V₀ * sin(θ) / g
t = 35 m/s * sin(60°) / 9.8 m/s²
t ≈ 3.05 seconds

Therefore, it will take around 3.05 seconds for the rock to reach the top of the arc (apogee).

Lastly, let's determine how far away the rock is from your hand when it is at the same level as your hand for the second time.

The horizontal distance traveled by the rock can be calculated using the horizontal component of the velocity (Vx) and the time taken (t) to reach the same level.

Distance = Vx * t
Distance = 17.5 m/s * 2t
Distance = 17.5 m/s * 2 * 3.05 s
Distance ≈ 106.15 meters

Therefore, the rock will be approximately 106.15 meters away from your hand when it is level with your hand for the second time.