Question

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Answer 1

Question -

An object is being thrown at a Speed of 20 m/s in a direction 45° above horizontal. The time taken by object to return to same level is -

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Answer

Given -

u = 20 m/s

$\theta$ = 45°

where

$\longrightarrow$u is initial velocity.

$\longrightarrow$$\theta$ is the angle of projection.

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To find -

Time to flight $\longrightarrow$ T

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Formula used -

$\boxed{\bf T = \frac{2u \sin\theta }{g}}$

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Derivation -

By 2nd equation of motion in y direction -

$\implies$$\bf S_y = u_yt + 1/2 gt^2$

$\longrightarrow$$\bf S_y = 0$

$\longrightarrow$$\bf u_y = -usin\theta$

$\longrightarrow$g is acceleration due to gravity.

$\longrightarrow$T is time taken to complete its motion $\longrightarrow$Time of flight.

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Substituting the value -

$\bf 0 = -usin\theta(T) + 1/2 g T^2$

$\implies$$\bf u sin\theta T = 1/2 g T^2$

$\implies$$\bf T = \frac{2u \sin\theta }{g}$

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Solution -

$\implies$u = 20 m/s

$\implies$$\theta$ = 45°

Substituting the value in the formula -

$\bf T = \frac{2u \sin\theta }{g}$

$\implies$$\bf = \frac{2 \times 20 \times \sin45}{g}$

$\implies$$\bf = \frac{2 \times 20 \times \frac{1}{ \sqrt{2} } }{g}$

$\implies$$\bf = \frac{20 \sqrt{2} }{g}$

So the time taken by object to complete its motion is $\bf \frac{20 \sqrt{2} }{g}$.

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Thanks

Answer 2

$\huge\pink{QuesTiOn}$

An object is being thrown at a Speed of 20 m/s in a direction 45° above horizontal. The time taken by object to return to same level is -

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$\huge\pink{AnsWeR}$

Given -

$u = 20 m/s$

$θ = 45°$

where

⟶ u is initial velocity.

⟶ θ is the angle of projection.

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To find -

Time to flight ⟶ T

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Formula used -

$/boxedT=<strong>2u</strong><strong> </strong>sinθ /g$

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Derivation -

By 2nd equation of motion in y direction -

$⟹ S_y = u_yt + 1/2 gt^2$

$⟶ \bf S_y = 0$

$⟶ u_y =−usinθ$

⟶ g is acceleration due to gravity.

⟶ T is time taken to complete its motion.

⟶ Time of flight.

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Substituting the value -

$0 =−usinθ(T)+1/2gT ^20$

$⟹ u sin\θ T =1/2gT ^2$

⟹ $T = \frac{2u \sin\theta }{g}$

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Solution -

$⟹ u = 20 m/s$

$⟹ θ = 45°$

Substituting the value in the formula -

⟹ $T = \frac{2u \sin\theta }{g}$

$⟹ \frac{2 \times 20 \times \sin45}{g}$

$⟹ \frac{2 \times 20 \times \frac{1}{ \sqrt{2} } }{g}$

$⟹ \frac{20 \sqrt{2} }{g}$

So the time taken by object to complete its motion is $\frac{20 \sqrt{2} }{g}$

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$<marquee>$

HOPE IT HELPS!!