If R is the maximum horizontal range of a particle, then the greatest height attained by it is: (1) R (2) 2R (3) R/2 (4) R/4. Two stones are projected with the same speed but makin... If R is the maximum horizontal range of a particle, then the greatest height attained by it is: (1) R (2) 2R (3) R/2 (4) R/4. Two stones are projected with the same speed but making different angles with the horizontal. Their ranges are equal. If the angle of projection of one is π/3 and its maximum height is y1, then the maximum height of the other will be: (1) 3y1 (2) 2y1 (3) y1/2 (4) y1/3. A projectile is thrown from a point in a horizontal plane such that its horizontal and vertical velocity components are 9.8 m/s and 19.6 m/s respectively. Its horizontal range is: (1) 4.9 m (2) 9.8 m (3) 19.6 m (4) 39.2 m.
Understand the Problem
The question is asking for the greatest height attained by a projectile in relation to its maximum horizontal range. It mentions two specific physics problems involving the projection of stones and the horizontal range of a projectile thrown from a horizontal plane.
Answer
The horizontal range is $19.6 \, \text{m}$.
Answer for screen readers
The horizontal range is $19.6 , \text{m}$.
Steps to Solve
- Understanding the problem for question 126
We need to find the horizontal range of a projectile given its horizontal and vertical velocity components.
- Using the horizontal range formula
The formula for the horizontal range $R$ of a projectile launched from an initial height is:
$$ R = \frac{v_x}{g} \cdot \sqrt{v_y^2 + 2gh} $$
Since there is no initial height, the simplified formula is:
$$ R = \frac{v_x}{g} \cdot v_y $$
- Plugging in the values
Given:
- Horizontal velocity component $v_x = 9.8 , \text{m/s}$
- Vertical velocity component $v_y = 19.6 , \text{m/s}$
- Acceleration due to gravity $g = 9.8 , \text{m/s}^2$
Substituting these values:
$$ R = \frac{9.8}{9.8} \cdot 19.6 $$
- Calculating the range
This simplifies to:
$$ R = 1 \cdot 19.6 $$
So:
$$ R = 19.6 , \text{m} $$
The horizontal range is $19.6 , \text{m}$.
More Information
The solution uses basic kinematics to find the horizontal range of a projectile, illustrating how both horizontal and vertical components of the initial velocity contribute to the distance traveled.
Tips
- Confusing horizontal and vertical components of the velocity.
- Forgetting to include the acceleration due to gravity in calculations.
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