Question:
The equation of path of a projectile moving in x-y plane is given by $ y = x - \frac{x^2}{25} $, then the initial speed and maximum height of particle will be (take $ g = 10 \, \text{m/s}^2 $):
The equation of path of a projectile moving in x-y plane is given by $ y = x - \frac{x^2}{25} $, then the initial speed and maximum height of particle will be (take $ g = 10 \, \text{m/s}^2 $):
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To find the maximum height quickly from a trajectory equation $ y = ax - bx^2 $, you can use calculus. The maximum height occurs at the vertex where $ dy/dx = 0 $.
Here, $ 1 - (2x/25) = 0 \implies x = 12.5 $. Plugging $ x = 12.5 $ into the equation gives $ y = 12.5 - (12.5)^2/25 = 12.5 - 6.25 = 6.25 $, which is exactly $ 25/4 $.
Updated On: May 12, 2026
- $ 4\sqrt{10} \, \text{m/s} $ and $ \frac{29}{4} \, \text{m} $
- $ 25\sqrt{10} \, \text{m/s} $ and $ \frac{21}{4} \, \text{m} $
- $ 2\sqrt{10} \, \text{m/s} $ and $ \frac{31}{4} \, \text{m} $
- $ 5\sqrt{10} \, \text{m/s} $ and $ \frac{25}{4} \, \text{m} $
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The Correct Option is D
Solution and Explanation
Concept:
The standard trajectory equation for a projectile launched with initial velocity $ u $ at an angle $ \theta $ with the horizontal is:
$$ y = x \tan \theta - \frac{gx^2}{2u^2 \cos^2 \theta} $$By comparing the given equation $ y = x - \frac{x^2}{25} $ with this standard form, we can extract the values for the launch angle and the initial velocity. Once these are found, the maximum height $ H_{max} $ can be calculated using the formula:$$ H_{max} = \frac{u^2 \sin^2 \theta}{2g} $$
Step 1: Comparing the equations to find the launch angle. Comparing $ y = x - \frac{x^2}{25} $ with $ y = x \tan \theta - \frac{gx^2}{2u^2 \cos^2 \theta} $: The coefficient of $ x $ gives: $$ \tan \theta = 1 \implies \theta = 45^\circ $$At $ \theta = 45^\circ $, we have:$$ \sin 45^\circ = \frac{1}{\sqrt{2}}, \quad \cos 45^\circ = \frac{1}{\sqrt{2}}, \quad \text{and} \quad \cos^2 45^\circ = \frac{1}{2} $$
Step 2: Finding the initial speed ($ u $). The coefficient of $ x^2 $ gives:$$ \frac{g}{2u^2 \cos^2 \theta} = \frac{1}{25} $$Substitute $ g = 10 $ and $ \cos^2 \theta = 1/2 $:$$ \frac{10}{2u^2 (1/2)} = \frac{1}{25} \implies \frac{10}{u^2} = \frac{1}{25} $$$$ u^2 = 250 \implies u = \sqrt{25 \times 10} = 5\sqrt{10} \, \text{m/s} $$
Step 3: Calculating the maximum height ($ H_{max} $). Now, substitute $ u^2 = 250 $ and $ \sin^2 \theta = 1/2 $ into the maximum height formula: $$ H_{max} = \frac{u^2 \sin^2 \theta}{2g} = \frac{250 \times (1/2)}{2 \times 10} $$$$ H_{max} = \frac{125}{20} $$Divide both by 5:$$ H_{max} = \frac{25}{4} \, \text{m} $$Thus, the initial speed is $ 5\sqrt{10} \, \text{m/s} $ and the maximum height is $ \frac{25}{4} \, \text{m} $.
Step 1: Comparing the equations to find the launch angle. Comparing $ y = x - \frac{x^2}{25} $ with $ y = x \tan \theta - \frac{gx^2}{2u^2 \cos^2 \theta} $: The coefficient of $ x $ gives: $$ \tan \theta = 1 \implies \theta = 45^\circ $$At $ \theta = 45^\circ $, we have:$$ \sin 45^\circ = \frac{1}{\sqrt{2}}, \quad \cos 45^\circ = \frac{1}{\sqrt{2}}, \quad \text{and} \quad \cos^2 45^\circ = \frac{1}{2} $$
Step 2: Finding the initial speed ($ u $). The coefficient of $ x^2 $ gives:$$ \frac{g}{2u^2 \cos^2 \theta} = \frac{1}{25} $$Substitute $ g = 10 $ and $ \cos^2 \theta = 1/2 $:$$ \frac{10}{2u^2 (1/2)} = \frac{1}{25} \implies \frac{10}{u^2} = \frac{1}{25} $$$$ u^2 = 250 \implies u = \sqrt{25 \times 10} = 5\sqrt{10} \, \text{m/s} $$
Step 3: Calculating the maximum height ($ H_{max} $). Now, substitute $ u^2 = 250 $ and $ \sin^2 \theta = 1/2 $ into the maximum height formula: $$ H_{max} = \frac{u^2 \sin^2 \theta}{2g} = \frac{250 \times (1/2)}{2 \times 10} $$$$ H_{max} = \frac{125}{20} $$Divide both by 5:$$ H_{max} = \frac{25}{4} \, \text{m} $$Thus, the initial speed is $ 5\sqrt{10} \, \text{m/s} $ and the maximum height is $ \frac{25}{4} \, \text{m} $.
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