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Standard XII

Physics

2nd Equation of Motion

A particle is...

Question

A particle is projected from horizontal making an angle $60^{\circ}$ with initial velocity 40 m/s. Find the time taken to the particle to make angle $45^{\circ}$ from horizontal.

2.5 s

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3.5 s

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1.5 s

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4.5 s

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Solution

The correct option is C $1.5 s$
In horizontal direction $v_{x} = u_{x}$
At particle makes $45^{\circ}, v_{x} = v_{y}$
Applying first equation of motion in y direction,
$v_{y} = u_{y} + a_{y} t$
or $u_{x} = u_{y} - g t$
$∴ t = \frac{u_{y} - u_{x}}{g} = \frac{40 (sin 60^{\circ} - sin 30^{\circ})}{9.8} \approx 1.5 s$

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A particle is projected from horizontal making an angle 60∘ with initial velocity 40 m/s. Find the time taken to the particle to make angle 45∘ from horizontal.

The correct option is C 1.5sIn horizontal direction vx=ux At particle makes 45∘,vx=vy Applying first equation of motion in y direction, vy=uy+ayt or ux=uy−gt ∴t=uy−uxg=40(sin 60∘−sin 30∘)9.8≈1.5s

A particle is projected from horizontal making an angle $60^{\circ}$ with initial velocity 40 m/s. Find the time taken to the particle to make angle $45^{\circ}$ from horizontal.