Math Problem Statement

. An object projected upwards reaches a maximum height of 100 m. When an object with a mass two-thirds that of the first object is thrown upwards with a speed twice the projectile speed of the first object, the maximum height it reaches is,

(1) 25 m

(2) 50 m

(3) 100 m

(4) 200 m

(5) 400 m

Solution

To solve this problem, we need to use the concept of energy conservation in projectile motion. The maximum height hh reached by an object in projectile motion can be derived using the principle of conservation of energy:

12mv2=mgh\frac{1}{2} m v^2 = mgh

Where:

  • mm is the mass of the object,
  • vv is the initial speed of the object,
  • gg is the acceleration due to gravity,
  • hh is the maximum height reached.

Step 1: Analyze the first object

For the first object, we know:

  • The maximum height is h1=100mh_1 = 100 \, \text{m}.
  • The initial speed is v1v_1.

Using the energy conservation equation for the first object: 12mv12=mgh1\frac{1}{2} m v_1^2 = m g h_1 v12=2gh1v_1^2 = 2 g h_1 v12=2g×100=200gv_1^2 = 2 g \times 100 = 200 g Thus, v12=200gv_1^2 = 200g.

Step 2: Analyze the second object

For the second object:

  • Its mass is 23\frac{2}{3} of the first object's mass.
  • Its initial speed is v2=2v1v_2 = 2 v_1.

We need to find the maximum height h2h_2 it reaches. Again, using energy conservation: 12×23m×v22=23mgh2\frac{1}{2} \times \frac{2}{3} m \times v_2^2 = \frac{2}{3} m g h_2 Substitute v2=2v1v_2 = 2 v_1: 12×23m×(2v1)2=23mgh2\frac{1}{2} \times \frac{2}{3} m \times (2 v_1)^2 = \frac{2}{3} m g h_2 12×23m×4v12=23mgh2\frac{1}{2} \times \frac{2}{3} m \times 4 v_1^2 = \frac{2}{3} m g h_2 Substitute v12=200gv_1^2 = 200 g: 12×23m×4×200g=23mgh2\frac{1}{2} \times \frac{2}{3} m \times 4 \times 200 g = \frac{2}{3} m g h_2 12×23m×800g=23mgh2\frac{1}{2} \times \frac{2}{3} m \times 800 g = \frac{2}{3} m g h_2 The 23m\frac{2}{3} m cancels out on both sides: 12×800g=gh2\frac{1}{2} \times 800 g = g h_2 400g=gh2400 g = g h_2 Thus, h2=400mh_2 = 400 \, \text{m}.

Conclusion:

The second object reaches a maximum height of 400 meters. Therefore, the correct answer is:

(5) 400 m

Would you like any more details or have any questions?

Here are 5 related questions to explore further:

  1. What would happen to the height if the initial speed was tripled instead of doubled?
  2. How does the mass of an object affect its maximum height if the speed is kept constant?
  3. Can we find the time taken to reach the maximum height for each object?
  4. What if the second object was thrown with the same speed as the first object?
  5. How would air resistance affect the outcome in real-world scenarios?

Tip: Doubling the initial speed of an object quadruples its maximum height in ideal conditions (without air resistance).

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Math Problem Analysis

Mathematical Concepts

Energy Conservation
Projectile Motion
Kinematics

Formulas

Conservation of energy: (1/2)mv^2 = mgh
Height formula: v^2 = 2gh

Theorems

Conservation of Mechanical Energy in Projectile Motion

Suitable Grade Level

Grades 10-12