The maximum height reached by a projectile is 64 m. If the initial velocity is halved, the new maximum height of the projectile is ______ m.
Correct Answer: 16
Approach Solution - 1
The maximum height of a projectile is given by:
\[ H_{\text{max}} = \frac{u^2 \sin^2 \theta}{2g}, \]
where:
- \( u \) is the initial velocity,
- \( \theta \) is the angle of projection,
- \( g \) is the acceleration due to gravity.
Step 1: Relationship between maximum heights
If the initial velocity is halved, i.e., \( u' = \frac{u}{2} \), the new maximum height \( H_{2\text{max}} \) can be expressed as:
\[ \frac{H_{1\text{max}}}{H_{2\text{max}}} = \frac{u^2}{u'^2}. \]
Substitute \( u' = \frac{u}{2} \):
\[ \frac{H_{1\text{max}}}{H_{2\text{max}}} = \frac{u^2}{\left( \frac{u}{2} \right)^2}. \]
Step 2: Simplify the expression
Simplify \( \left( \frac{u}{2} \right)^2 \):
\[ \frac{H_{1\text{max}}}{H_{2\text{max}}} = \frac{u^2}{\frac{u^2}{4}} = 4. \]
Thus:
\[ H_{2\text{max}} = \frac{H_{1\text{max}}}{4}. \]
Step 3: Calculate the new maximum height
Substitute \( H_{1\text{max}} = 64 \, \text{m} \):
\[ H_{2\text{max}} = \frac{64}{4} = 16 \, \text{m}. \]
Therefore, the new maximum height of the projectile is \( H_{2\text{max}} = 16 \, \text{m} \).
Approach Solution -2
Initial maximum height, H₁ = 64 m
Initial velocity = u
New velocity = u/2
Step 2: Formula for maximum height of a projectile.
The maximum height reached by a projectile is given by:
\[ H = \frac{u^2 \sin^2 \theta}{2g} \]
Step 3: Relation between heights and velocities.
Since \( H \propto u^2 \), if the velocity is halved, the new height H₂ will be:
\[ \frac{H_2}{H_1} = \left( \frac{u/2}{u} \right)^2 = \left( \frac{1}{2} \right)^2 = \frac{1}{4} \] \[ H_2 = \frac{H_1}{4} = \frac{64}{4} = 16 \, \text{m} \]
Step 4: Final Answer.
The new maximum height of the projectile is:
\[ \boxed{16 \, \text{m}} \]
Final Answer: 16 m
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