An object of mass is projected with a momentum at an angle with the horizontal such that its maximum height (H) is half of its Range (R). Minimum kinetic energy of the particle in its path will be

Question

An object of mass $m$ is projected with a momentum $p$ at an angle $θ$ with the horizontal such that its maximum height (H) is half of its Range (R). Minimum kinetic energy of the particle in its path will be

Smart Achievers · Accepted Answer

(3)

$1 \times \frac{1}{2 g} {(\frac{p \sin θ}{m})}^{2} = \frac{p \sin θ}{m g} \times \frac{p \cos θ}{m}$

$\frac{1}{2} \sin^{2} θ = \sin θ \cos θ \Rightarrow \tan θ = 2$

$∴$ $\cos θ = \frac{1}{\sqrt{5}}$

$Minimum kinetic energy = {(\frac{p \cos θ}{2 m})}^{2} = \frac{p^{2}}{2 m} \times \frac{1}{5} = \frac{p^{2}}{10 m}$

Solution

Q.
An object of mass $m$ is projected with a momentum $p$ at an angle $θ$ with the horizontal such that its maximum height (H) is half of its Range (R). Minimum kinetic energy of the particle in its path will be

1 $\frac{p^{2}}{8 m}$

2 $\frac{3 p^{2}}{4 m}$

3 $\frac{p^{2}}{10 m}$

4 $\frac{p^{2}}{5 m}$

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Solution

Q. An object of mass m is projected with a momentum p at an angle θ with the horizontal such that its maximum height (H) is half of its Range (R). Minimum kinetic energy of the particle in its path will be

1 p28m

2 3p24m

3 p210m

4 p25m