2. An object of mass 200 g (0.2 kg) is thrown vertically upwards with a velocity of 8 m/s from a height of 10 m above the ground. It reaches certain maximum height and falls straight back to ground. Ignore air resistance in this problem. (a) Draw v-t graph that represents the velocity of the object after it is launched upwards and until it hits the ground. (b) How long does it take for the object to reach the maximum height? (c) What is the maximum height of the object from the ground? (d) How long does it take for the object to reach the ground from the highest point of motion? (e) What is the velocity of the object as it reaches ground? (Hint: Not zero).

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### Physics Problem: Vertically Thrown Object

**Problem Statement:**
An object of mass 200 g (0.2 kg) is thrown vertically upwards with a velocity of 8 m/s from a height of 10 m above the ground. It reaches a certain maximum height and falls straight back to the ground. Ignore air resistance in this problem.

1. Draw a velocity-time (v-t) graph that represents the velocity of the object after it is launched upwards and until it hits the ground.
2. How long does it take for the object to reach the maximum height?
3. What is the maximum height of the object from the ground?
4. How long does it take for the object to reach the ground from the highest point of motion?
5. What is the velocity of the object as it reaches the ground? (Hint: Not zero).

**Explanation:**

(a) **v-t Graph**:
   - The velocity-time graph for this motion will be a straight line with a negative slope due to the constant acceleration of gravity. The velocity starts at +8 m/s (upwards) and decreases linearly to 0 m/s at the maximum height, then becomes negative and increases in magnitude until the object hits the ground.

(b) **Time to Reach Maximum Height**:
   - The time to reach maximum height can be calculated using the formula \(v = u + at\), where \(u\) is the initial velocity (8 m/s), \(v\) is the final velocity (0 m/s at maximum height), and \(a\) is the acceleration (-9.8 m/s²). Solving for \(t\), we get:
     \[
     0 = 8 + (-9.8)t \implies t = \frac{8}{9.8} \approx 0.82 \text{ seconds}
     \]

(c) **Maximum Height from the Ground**:
   - The maximum height can be found using the kinematic equation \(v^2 = u^2 + 2as\), where \(s\) is the displacement. Solving for \(s\):
     \[
     0 = 8^2 + 2(-9.8)s \implies s = \frac{64}{19.6} \approx 3.27 \text{ meters}
     \]
   - Adding the initial height of 10 meters, the total maximum height from the
Transcribed Image Text:### Physics Problem: Vertically Thrown Object **Problem Statement:** An object of mass 200 g (0.2 kg) is thrown vertically upwards with a velocity of 8 m/s from a height of 10 m above the ground. It reaches a certain maximum height and falls straight back to the ground. Ignore air resistance in this problem. 1. Draw a velocity-time (v-t) graph that represents the velocity of the object after it is launched upwards and until it hits the ground. 2. How long does it take for the object to reach the maximum height? 3. What is the maximum height of the object from the ground? 4. How long does it take for the object to reach the ground from the highest point of motion? 5. What is the velocity of the object as it reaches the ground? (Hint: Not zero). **Explanation:** (a) **v-t Graph**: - The velocity-time graph for this motion will be a straight line with a negative slope due to the constant acceleration of gravity. The velocity starts at +8 m/s (upwards) and decreases linearly to 0 m/s at the maximum height, then becomes negative and increases in magnitude until the object hits the ground. (b) **Time to Reach Maximum Height**: - The time to reach maximum height can be calculated using the formula \(v = u + at\), where \(u\) is the initial velocity (8 m/s), \(v\) is the final velocity (0 m/s at maximum height), and \(a\) is the acceleration (-9.8 m/s²). Solving for \(t\), we get: \[ 0 = 8 + (-9.8)t \implies t = \frac{8}{9.8} \approx 0.82 \text{ seconds} \] (c) **Maximum Height from the Ground**: - The maximum height can be found using the kinematic equation \(v^2 = u^2 + 2as\), where \(s\) is the displacement. Solving for \(s\): \[ 0 = 8^2 + 2(-9.8)s \implies s = \frac{64}{19.6} \approx 3.27 \text{ meters} \] - Adding the initial height of 10 meters, the total maximum height from the
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