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-

At 45^∘, v_x=v_y or u_x=u_y gt ∴ t=u_y u_x/g=40(sin 60^∘ sin 30^∘)/9.8=1.5s ...

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Byju's Answer

Standard XII

Physics

Intro to Projectile on an Incline

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.
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Solution

At $45^{\circ}$ , $v_{x} = v_{y}$
or $u_{x} = u_{y} - g t$
$∴ t = \frac{u_{y} - u_{x}}{g}$ $= \frac{40 (s i n 60^{\circ} - s i n 30^{\circ})}{9.8}$ =1.5s

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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. ___

At 45∘, vx=vy 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.
___

At $45^{\circ}$ , $v_{x} = v_{y}$
or $u_{x} = u_{y} - g t$
$∴ t = \frac{u_{y} - u_{x}}{g}$ $= \frac{40 (s i n 60^{\circ} - s i n 30^{\circ})}{9.8}$ =1.5s