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Need explanation for: - Kinematics - JEE Main-2

A shell is fired from a fixed artillery gun with an initial speed u such that it hits the target on the ground at a distance R from it. If  t_{1} and t_{2} are the values of the time taken by it to hit the target in two possible ways, the product t_{1}t_{2} is:

  • Option 1)

    R/2g

  • Option 2)

    R/4g

  • Option 3)

    R/g

  • Option 4)

    2R/g

 
Answers (1)
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Horizontal Range -

Horizontal distance travelled by projectile from the point of projectile to the point on ground where its hits.

R=\frac{u^{2}\sin 2\Theta }{g}

 

 

 

- wherein

Special case of horizontal range

For max horizontal range.

\Theta = 45^{0}

R_{max}=\frac{u^{2}\sin 2 (45) }{g}=\frac{u^{2}\times 1}{g}=\frac{u^{2}}{g}

 

 

If the range is equal for both the target then the angle of projection may be different.

t_{1}=\frac{2u}{g}\sin\theta_{1}  ;   t_{2}=\frac{2u\sin\theta_{2}}{g}

\Rightarrow t_{1}t_{2}=\frac{4u^2}{g^2}\sin\theta_{1}\sin\theta_{2}\because ranges are  equal.

\theta_{2}=90^{\circ}-\theta_{1} \therefore t_{1}t_{2}=\frac{2}{g}\frac{2u^2\sin\theta_{1}\cos\theta_{1}}{g}=\frac{2R}{g}


Option 1)

R/2g

Option 2)

R/4g

Option 3)

R/g

Option 4)

2R/g

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