HRWP49p79

In uniform circular motion, the magnitude of the acceleration is v²/r. Since v = 2π r/T then a = 4π² r/T². Now we see that for any object on the merry-go-round, the magnitude of the acceleration depends linearly on r. Since multiplying a vector by a scalar does not change the direction of the vector, objects along a single radial line will have accelerations which only differ by a scalar quantity.

HRWP36p78

To solve this problem, you have to do a bunch of steps. I'm listing the steps to part a in reverse order since that is how you would need to tackle the problem:

1. To find d, you need the horizontal speed (v_x) since
   d = (v_x) t and t=4s.


2. To find v_x, you need the total velocity (v) when it lands on
   the roof (because v_x = v cos(60°))


3. To find v, you need to find the vertical velocity (v_y) when it
   lands on the roof.


4. To find the final vertical velocity (v_y), you need the initial
   vertical velocity (v_y0).


5. To find v_y0, use the acceleration equation:
   
   h = .5 g t² + v_y0 t


6. Now, you can find v_y since v_y = v_y0 + gt


7. Next, you can find v since v_y = v sin(60°)


8. Going backwards, you can find d.


9. Parts b and c are easy since you already have v_y0 and
   v_x (it's the usual translation from cartesian to polar coordinates).
   

HRWP33p78

1. Using the distance and time, you can find the horizontal velocity (which is assumed to be constant).
2. Using the angle of impact, you can find the final vertical velocity.
3. "Running the tape backwards", you can use the usual vertical equation along with the time to get h.
4. Using the time and the final vertical velocity, you can get the initial vertical velocity (still running the tape backwards).
5. You can now get the total initial velocity and the initial angle.

HRWP31p78

1. Assume an angle, theta.
2. Split the initial velocity into vertical and horizontal components (sine and cosine).
3. Write down the horizontal equation (using the V_x from step 2).
4. Write down the usual vertical equation but h = 0 (bullet starts and ends at same height).
5. Solve horizontal equation for t.
6. Plug t into vertical equation, use some trig identities, and solve for theta.
7. Finally, solve for distance above target (more simple trig).

HRWP28p78

The time the ball is in the air depends on the initial vertical component of the velocity and on gravity so you have to split the initial velocity into vertical and horizontal components. The vertical gives you the time in the air and the horizontal tells you how far away it lands. That gives you the soccer player's required distance and you already have the time.

HRWP22p77

You have the initial velocity. Use sine and cosine to break it into components (vx (ihat) + vy (jhat)). Now, you can ignore the horizontal part and just the vertical velocity to find the height at 5.5s (part a). You can also find the vertical (negative) velocity at that time. Combine with the constant horizontal velocity to get the total velocity at that time (part b). Finally, you can find out how long it takes to reach it's maximum height (zero velocity) and calculate how high it will be (a vertical-only equation but watch out for signs).

HRWP19p77

You start with an initial height (think position vector) and an initial velocity vector. You have other information, such as the acceleration due to gravity of g = -9.8 m/s^2 (jhat). You can solve the vertical equation using just the (jhat) components of the acceleration and initial velocity (and initial height). See the second equation on your AP formula sheet (mechanics) if you need a reminder of how this goes. After part a, figure out how long till the ball hits the ground and you can solve the horizontal equation (the (ihat) part).

HRWP18p77

See formula 4-26 on page 67.

HRWP17p77

This is just the cannonball lab from last year. You have the second equation on your equation sheet for the vertical motion and the time will give you the horizontal distance.

HRWP11p77

For part a you just plug in 2.00s for t

For part b, you calculate the velocity vector and plug in 2.00s for t, etc.

HRWP9p77

See the post, Dr. G's Physics Blog: HRWP6p77.

HRWP6p77

The velocity vector is the time derivative of the position vector. For this type of derivative, the unit vectors (such as (ihat)) are constants. Only the t parts get their derivatives taken. Thus if r = (5t, 3) m, v = (5, 0) m/s since the derivative of 5t is 5 and the derivative of 3 is zero.
The acceleration vector is derived from the velocity vector in the same manner.