Question

a ball is projected upward with an initial speed of approximately 30 m/s. the diagram at the right represents its position at 1-second intervals of time. at what location will the ball be moving downward with a speed of approximately 30 m/s? tap to toggle through answer options: tap here

Explanation:

Step1: Recall Free Fall Motion

In free fall (or projectile motion with only gravity), the acceleration due to gravity is constant ($g \approx 10 \, \text{m/s}^2$). The speed of an object projected upward changes by $g$ each second. When projected upward with initial speed $v_0 = 30 \, \text{m/s}$, it slows down by $10 \, \text{m/s}$ each second until it reaches the peak (speed = 0), then speeds up downward by $10 \, \text{m/s}$ each second.

Step2: Analyze Speed Change

- At $t = 0$ (launch), speed is $30 \, \text{m/s}$ upward.
- At $t = 1 \, \text{s}$, speed is $30 - 10 = 20 \, \text{m/s}$ upward.
- At $t = 2 \, \text{s}$, speed is $20 - 10 = 10 \, \text{m/s}$ upward.
- At $t = 3 \, \text{s}$, speed is $10 - 10 = 0 \, \text{m/s}$ (peak).
- Then, it starts falling:
- At $t = 4 \, \text{s}$, speed is $0 + 10 = 10 \, \text{m/s}$ downward.
- At $t = 5 \, \text{s}$, speed is $10 + 10 = 20 \, \text{m/s}$ downward.
- At $t = 6 \, \text{s}$, speed is $20 + 10 = 30 \, \text{m/s}$ downward.

Step3: Relate to Position (Symmetry)

The motion is symmetric: the time to reach the peak (where speed is 0) and the time to fall back to the same speed (magnitude) downward are equal. Since it took 3 seconds to reach the peak (from $v_0 = 30 \, \text{m/s}$ upward to $v = 0$), it will take another 3 seconds to fall back to a speed of $30 \, \text{m/s}$ downward.

Assuming the grid represents 1-second intervals, the location corresponding to $t = 6 \, \text{s}$ (or the symmetric point below the launch height, at the same vertical distance but moving downward) would be the location where speed is $30 \, \text{m/s}$ downward. From the diagram, the point symmetric to the launch point (A/G? Wait, no—wait, the initial launch is A? Wait, the diagram has A, G; B, F; C, E; D. Wait, the vertical spacing: each row is 1 second? Wait, the key is symmetry: the speed upward at launch (30 m/s) will be equal in magnitude (30 m/s) downward when it has fallen back to the same vertical position (or the time when the total change in speed is $60 \, \text{m/s}$? No, wait: initial speed $v_0 = 30 \, \text{m/s}$ upward. At peak, speed is 0. Then, to get speed $30 \, \text{m/s}$ downward, the change in speed from peak is $30 \, \text{m/s}$, which takes $t = \frac{30}{10} = 3 \, \text{s}$ after peak. Since peak is at $t = 3 \, \text{s}$ (after launch), then $t = 3 + 3 = 6 \, \text{s}$ after launch.

Looking at the diagram, the points: A and G are at the bottom, B and F, C and E, D at the top. The symmetric point to the launch (A) would be G? Wait, no—wait, the vertical distance: each interval is 1 second. The launch is at A (or G? Wait, the initial projection is upward, so A is launch, then moving up: B, C, D (peak at D), then down: C, E, B, F, A, G? Wait, no, the time intervals: 1-second intervals. So from A (t=0) up to D (t=3, peak), then down: D (t=3), C (t=4), E (t=5), B (t=6), F (t=7), A (t=8), G (t=9)? No, maybe the vertical grid is time. Wait, the key is that the speed downward of 30 m/s occurs at the same vertical position as the launch but on the way down, or symmetric in time. Since it takes 3 seconds to go up to peak (speed 0), then 3 seconds to come down to speed 30 m/s downward (since from peak, speed increases by 10 m/s each second: 10 at t=4, 20 at t=5, 30 at t=6). So at t=6, the position is F (or B? Wait, the diagram: B and F are at the same vertical level? Wait, the points: A and G (bottom), B and F (next), C and E (next), D (top). So when moving down, at t=6, the position is F (if B is t=1, F is t=7? No, maybe I got the time wrong. Wait, initial speed 30 m/s upward. Time to peak: $v = v_0 - gt \implies 0 = 30 - 10t \implies t = 3 \, \text{s}$. So peak at t=3 (D). Then, at t=4: speed 10 m/s downward (position C), t=5: 20 m/s downward (position E), t=6: 30 m/s downward (position B? No, wait, t=6: from t=3 (D) to t=6 is 3 seconds, so speed is $v = 0 + 10(3) = 30 \, \text{m/s}$ downward. The position at t=6: since it took 3 seconds to go up from A to D, it will take 3 seconds to go down from D to the same vertical position as A (since symmetric). Wait, no—wait, the vertical displacement: $y = v_0 t - \frac{1}{2}gt^2$. At t=0, y=0 (A). At t=6, $y = 30(6) - 5(36) = 180 - 180 = 0$. So at t=6, it's back to the same vertical position as launch (A), but moving downward. Wait, but the diagram has A and G. Maybe G is at t=6? Wait, the problem says "the diagram at the right represents its position at 1-second intervals of time". So each dot is 1 second apart. So A (t=0), B (t=1), C (t=2), D (t=3, peak), C (t=4), E (t=5), B (t=6), F (t=7), A (t=8), G (t=9)? No, that can't be. Wait, maybe the vertical axis is position, and horizontal is time? No, the diagram is a grid with points: D at the top, C and E in the middle, B and F, A and G at the bottom. So the vertical distance between A and B is 1 second, B and C is 1 second, C and D is 1 second (peak at D), then D to C (1s), C to E (1s), E to B (1s), B to F (1s), F to A (1s), A to G (1s)? No, this is confusing. But the key physics concept: in free fall, the speed upward at launch (30 m/s) will be equal in magnitude (30 m/s) downward when the object has fallen back to the same vertical position (since the time to go up to peak is equal to the time to come down from peak to the launch height, and speed at launch upward is equal to speed at launch height downward). Wait, no—actually, when the object returns to the launch height, its speed is 30 m/s downward (due to symmetry: the motion is symmetric, so speed at same height is equal in magnitude, opposite in direction). So the location where it's moving downward with 30 m/s is the same vertical position as the launch (A), but on the way down. Looking at the diagram, A and G: maybe G is the position when it's moving downward at 30 m/s. Wait, but the options: the points are A, B, C, D, E, F, G. The correct position is G (or A? Wait, no—when it's moving downward at 30 m/s, it's passing through the launch height (A) with speed 30 m/s downward. But in the diagram, A and G are at the bottom. Wait, maybe the initial launch is at G, moving up to D, then down to A, then to G? No, this is unclear, but the physics principle is: the speed downward of 30 m/s occurs at the same vertical position as the launch (where speed was 30 m/s upward) but on the way down, due to symmetry. So the location is G (or A, but likely G as the next point after A). Wait, the problem's diagram: A and G are at the bottom, B and F, C and E, D at the top. So when moving down, after A (launch height), it goes to G. So the answer is G (or A? Wait, no—wait, the initial speed is 30 m/s upward at A. When it comes back to A, speed is 30 m/s downward. But in the diagram, A and G: maybe G is beyond A. Wait, the key is that the time to reach 30 m/s downward is 6 seconds after launch (as calculated), and the position at 6 seconds is F? No, I think I made a mistake. Let's re-express:

Initial speed $v_0 = 30 \, \text{m/s}$ upward.

Acceleration $a = -g = -10 \, \text{m/s}^2$ (upward is positive).

Speed as a function of time: $v(t) = v_0 + at = 30 - 10t$.

We want $v(t) = -30 \, \text{m/s}$ (downward, so negative).

Set $-30 = 30 - 10t$:

$-30 - 30 = -10t$

$-60 = -10t$

$t = 6 \, \text{s}$.

So at $t = 6 \, \text{s}$, the speed is 30 m/s downward. Now, the diagram: each dot is 1-second interval. So t=0: A, t=1: B, t=2: C, t=3: D (peak), t=4: C, t=5: E, t=6: B, t=7: F, t=8: A, t=9: G. Wait, no, that can't be. Alternatively, the vertical axis is position, and horizontal is time. So D is at t=3 (peak), C at t=2 and t=4, E at t=1 and t=5, B at t=0 and t=6, F at t=-1 and t=7, A at t=-2 and t=8, G at t=-3 and t=9. No, this is not helpful. But the key is that at t=6, the position is B (since t=0 is B? No, the problem says "the diagram at the right represents its position at 1-second intervals of time". So the first position (launch) is A, then after 1s: B, 2s: C, 3s: D (peak), 4s: C, 5s: E, 6s: B, 7s: F, 8s: A, 9s: G. So at t=6s, the position is B? No, that contradicts. Wait, no—when moving up, from A (t=0) to B (t=1), C (t=2), D (t=3, peak). Then down: D (t=3), C (t=4), E (t=5), B (t=6), F (t=7), A (t=8), G (t=9). So at t=6, position is B. But speed at t=6: v(6) = 30 - 10*6 = -30 m/s (downward), so yes, at position B? No, this is confusing. But the correct physics answer: the speed downward of 30 m/s occurs at the same vertical position as the launch (A) but on the way down, or at the position symmetric to the launch in time. Since the time to go up to peak is 3s, the time to come down to speed 30 m/s downward is 3s after peak, so 6s after launch. The position at 6s is the same as the position at 0s (launch) but moving downward, so the location is G (or A, but in the diagram, G is below A). Wait, maybe the diagram has A and G as the lowest points, so when it's moving downward at 30 m/s, it's at G. But I think the correct position is G (or the point symmetric to the launch point, which is G).

Answer:

G (or the position corresponding to 6 seconds after launch, which is G in the diagram)