Physics AP12d Kinematics

Question

 An air-to-ground rocket is fired horizontally from an A-10 attack aircraft flying at 200 m/s. Immediately after leaving the jet, the rocket accelerates horizontally at a rate of 50.0 m.s^2 as it falls vertically at 9.80 m/s2 for a time interval of 10.0 s.

a. How far does the rocket travel horizontally (relative to the plane) during the 10.0s?

b. How far does the rocket travel vertically during the 10.0 s?

c. What total displacement did the rocket travel (relative to the plane) in the 10.0 s?

d. What is the magnitude of the rocket’s velocity vector (relative to the plane) at the end of 10.0 s?

e. What is the direction of the rocket’s velocity vector (relative to the plane) at the end of the 10.0s (expressed as degrees below the horizontal)?

f. What is the magnitude of the rocket’s velocity vector relative to the ground at the end of 10.0 s? (Assume the plane)

g. What is the direction of the rocket’s velocity vector relative to the ground at the end of the 10.0s (expressed as degrees below the horizontal)?

Answer 1

To do this problem, consider what happens in the x-direction (horizontal) and what happens in the y-direction (vertical) as separate things.

a) Calculate displacement in the x-direction for the constant acceleration (of 50 m/s²) using:

x(t) = (1/2)a_xt² + v_i-xt + x₀ where a_x = 50.0 m/s², v_i-x = 0 m/s since we are calculating displacement relative to the plane and the rocket is initially traveling at the same speed as the plane, x₀ = 0 meters since we are calculating displacement from the initial firing position and calling that initial firing position our "zero" position, and t = 10 s.

b) Calculate displacement in the y-direction for the constant acceleration (of -9.8 m/s²) using:

y(t) = (1/2)a_yt² + v_i-yt + y₀ where a_y = -9.80 m/s², v_i-y = 0 m/s since there is no vertical motion initially, y₀ = 0 meters since we are calculating displacement from the initial firing position and calling that initial firing position our "zero" position, and t = 10 s.

c) Total displacement is calculated using the Pythagorean Theorem: d = √(x² + y²) where "x" is the distance calculated in "a)" and "y" is the distance calculated in "b)".

d) Calculate the velocity in the x-direction from:

v_x-f = a_xt + v_i-x where v_x-f is the final velocity (after 10 seconds) in the x-direction, a_x is again 50.0 m/s², t = 10.0 s, and v_i-x is again 0 m/s (since again we are measuring relative to the aircraft).

Calculate the velocity in the y-direction from:

v_y-f = a_yt + v_i-y where v_y-f is the final velocity (after 10 seconds) in the y-direction, a_y is again -9.80 m/s², t = 10.0 s, and v_i-y is again 0 m/s (since again there is no initial vertical velocity).

Then combine the vx-f and vy-f into a single velocity magnitude using the Pythagorean Theorem: v_f = √(v_x-f² + v_y-f²)

e) Using the v_x-f and v_y-f from "d)", calculate the direction using the trig tangent ratio:

θ = tan^-1( v_y-f/v_x-f)

f) and g) Add 200 m/s to v_x-f that was calculated in "d)" and redo the magnitude and direction calculations you did above.