Problem #2: An apple is thrown upward with an initial velocity of +24.0 m/s.

Question

a. Sketch the apple’s trip and label what you know.

b. How high will the apple go? (Ans: +29.4 m)

c. What is the total time needed for the apple to return to its original position? (Ans: 4.90 s)

USING THE GUESS METHOD SHOW EVERY STEP

Answer 1

I can't draw or insert an image here, so I will describe a.

a) Apple being tossed upwards with v₀ (or v_i) of +24.0 m/s, if you draw the full path it will be a straight line up and down, but is usually displayed as a very narrow parabola.

b) There are a handful of kinematic equations that can be used to solve this one, including combinations of them, but the most straightforward one is v_f²=v_i²+2*a*d. We were given the initial velocity, know that its final speed (at the peak) will be 0 m/s, and that the acceleration due to gravity is -9.81 m/s². Plugging values into the equation:

1. Plug in known values: (0 m/s)² = (24 m/s)² + 2 * (-9.81m/s²) * d
2. Do the squaring and multiplying: 0 m²/s² = 576 m²/s² + -19.62 m/s² * d
3. Move the 576 to the left side (this also combines things with similar units): -576 m²/s² = -19.62 m/s² * d
4. Divide both sides by -19.62 m/s²: (-576 m²/s² / -19.62 m/s²) = d
5. Do the division: 29.4 m = d

c) For the total time y=v_avg*t and average velocity is v_avg=(v_i+v_f)/2

1. Plug in known values: 29.4 m = ((24+0)m/s)/2 *t
2. Do the division: 29.4 m = 12 m/s * t
3. Divide both sides by 12 m/s: 29.4 m/(12 m/s) = t
4. Do the division: 2.45 s = t [m divided by m/s will result in the meters cancelling out leaving 1/1/s which rearranges to s]
5. This is the time to get to the peak, so double it for the full path: 2.45 s *2 = 4.90 s.

As for guess method, I don't know exactly what they are looking for, but a quick estimate in my head is that 24 m/s up will lose 10 m/s each second and so it will take 2.4 seconds to get to the peak where it has no speed, and double this to get 4.8 seconds. Then from there the average speed is 12 m/s [(24+0)/2 = 12] and it traveled for 2.4 seconds to the peak means 12 *2.4 for the maximum height [or quick guesstimate in my head is 12 * 2.5 which means double 12 and add half of 12 to it, so 24 + 6 = 30 m].