Im not sure how to answer this question

Question

Suppose a large rock is ejected from the volcano with a speed of 27m/s and at an angle of 33 degrees above the horizontal, as shown in the figure below. The rock strikes the ground at an altitude of 20m lower than its starting point. Include units as appropriate below. Determine the time it takes the rock to follow this path. What is the speed and angle below the positive horizontal which the rock impacts the side of the volcano?

Answer 1

Calculate the vertical component of the rock leaving the volcano (V_y-i) using the trig ratio sine:

sin(θ) = opposite/hypotenuse

sin(θ) = V_i-y/V_i

sin(33°) = V_y-i/27

V_y-i = 27sin(33°) m/s (use your calculator to get the value)

Calculate the height at which the rock reaches its maximum point (where the vertical velocity is zero) using:

v_f² = v_i² + 2a(y)

where v_f = zero, v_i = V_y-i calculated above, a = g = -9.81 m/s², and "y" is the height of the rock.

Now, add 20 meters to this to get the vertical distance the rock will fall.

Now, using the same equation v_f² = v_i² + 2a(y) for the fall of the rock to calculate the vertical component of the velocity when it hits the ground (V_f-y) where vf = V_f-y, v_i = zero, a = g = 9.81 m/s² (use a positive to make the signs work out. Keep in mind the velocity you calculate will be in the downward direction), and y = The height (including the 20 meters) that you calculated.

Now calculate the x-component of the original velocity of the rock leaving the volcano. This velocity will remain the same throughout its flight. To do so, use the trig ratio cosine:

cos(θ) = adjacent/hypotenuse

cos(θ) = V_i-x/V_i

cos(33°) = V_i-x/27

V_i-x = 27cos(33°) m/s (use your calculator to get the value)

V_f-x = V_i-x (the velocity in the x-direction has no force applied to it to change its speed so it remains the same)

Now you have both components of the velocity and can put them together to answer the question:

Use the Pythagorean Theorem to put them back together into the speed:

speed = √[(V_f-x)² + (V_f-y)²]

Use the trig ratio tangent to calculate the angle:

tan(θ) = opposite/adjacent

tan(θ) = V_f-y/V_f-x

θ = tan^-1(V_f-y/V_f-x)