Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

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)

best_answer

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)

$2R/g$

Posted by

solutionqc

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

What's the minimum marks to qualify JEE Main 2025? NTA score vs percentile for new paper pattern

anam-khan

Challenge JEE Main 2025 answer key for session 1 by today

anam-khan

JEE Main 2025 final answer key likely in February 2nd week; result date, marking scheme

Latest Question