A projectile is fired from horizontal ground with speed and projection angle . When the acceleration due to gravity is , the range of the projectile is . If at the highest point in its trajectory, the projectile enters a different region where the effect

Question

A projectile is fired from horizontal ground with speed $v$ and projection angle $θ$ . When the acceleration due to gravity is $g$ , the range of the projectile is $d$ . If at the highest point in its trajectory, the projectile enters a different region where the effective acceleration due to gravity is $g' = \frac{g}{0.81}$ , then the new range is $d' = n d$ . The value of $n$ is ____ . [2022]

Smart Achievers · Accepted Answer

(0.95)

After entering in new region, time taken by projectile to reach ground is given as

$t = \sqrt{\frac{2 h}{g_{eff}}} = \sqrt{\frac{2 \times 0.81 \times u^{2} \sin^{2} θ}{g \times 2 g}} = 0.9 \frac{u \sin θ}{g}$

So, horizontal displacement done by projectile in new region is given as

$x = 0.9 \times \frac{u \sin θ}{g} \times u \cos θ = 0.9 (\frac{d}{2})$

$Now, new range = \frac{d}{2} + 0.9 \frac{d}{2} = 0.95 d$

Solution

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Solution

Ans. (0.95) After entering in new region, time taken by projectile to reach ground is given as t=2hgeff=2×0.81×u2sin2θg×2g=0.9usinθg So, horizontal displacement done by projectile in new region is given as x=0.9×usinθg×ucosθ=0.9(d2) Now, new range =d2+0.9d2=0.95d

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