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� = � �.�� , then the new range is 𝑑� = 𝑛𝑑. The value of n is _____.

 We are given a projectile fired from horizontal ground with speed 

$v$ and projection angle $\theta$. The range of the projectile under gravity $g$ is $d$. At the highest point of its trajectory, the projectile enters a region where the effective acceleration due to gravity changes to $g^{\mathrm{\prime }}=\frac{g}{1.25}$. We are to find the factor $n$ such that the new range $d^{\mathrm{\prime }}=nd$.

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Step 1: Range of the projectile under gravity $g$

The range $d$ of a projectile under gravity $g$ is given by:

$$d=\frac{v^2\sin (2\theta )}{g}\mathrm{.}$$

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Step 2: Time of flight under gravity $g$

The time of flight $T$ of the projectile under gravity $g$ is:

$$T=\frac{2v\sin \theta }{g}\mathrm{.}$$

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Step 3: Time to reach the highest point

The time $t_1$ to reach the highest point is half the total time of flight:

$$t_1=\frac{T}{2}=\frac{v\sin \theta }{g}\mathrm{.}$$

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Step 4: Horizontal distance covered before the highest point

The horizontal distance $d_1$ covered before the highest point is:

$$d_1=v\cos \theta \cdot t_1=v\cos \theta \cdot \frac{v\sin \theta }{g}=\frac{v^2\sin \theta \cos \theta }{g}\mathrm{.}$$

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Step 5: Time of flight after the highest point

After the highest point, the projectile enters a region with gravity $g^{\mathrm{\prime }}=\frac{g}{1.25}$. The time $t_2$ to fall from the highest point to the ground under gravity $g^{\mathrm{\prime }}$ is:

$$t_2=\sqrt{\frac{2h}{g^{\mathrm{\prime }}}},$$

where $h$ is the maximum height of the projectile. The maximum height $h$ is:

$$h=\frac{v^2{\sin }^2\theta }{2g}\mathrm{.}$$

Substitute $h$ and $g^{\mathrm{\prime }}=\frac{g}{1.25}$ into $t_2$:

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Step 6: Horizontal distance covered after the highest point

The horizontal distance $d_2$ covered after the highest point is:

$$d_2=v\cos \theta \cdot t_2=v\cos \theta \cdot \frac{\sqrt{1.25}v\sin \theta }{g}=\frac{\sqrt{1.25}{v}^2\sin \theta \cos \theta }{g}\mathrm{.}$$

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Step 7: Total new range $d^{\mathrm{\prime }}$

The total new range $d^{\mathrm{\prime }}$ is the sum of $d_1$ and $d_2$:

$$d^{\mathrm{\prime }}=d_1+d_2=\frac{v^2\sin \theta \cos \theta }{g}+\frac{\sqrt{1.25}{v}^2\sin \theta \cos \theta }{g}\mathrm{.}$$

Factor out $\frac{v^2\sin \theta \cos \theta }{g}$:

$$d^{\mathrm{\prime }}=\frac{v^2\sin \theta \cos \theta }{g}(1+\sqrt{1.25})\mathrm{.}$$

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Step 8: Compare with the original range $d$

The original range $d$ is:

$$d=\frac{v^2\sin (2\theta )}{g}=\frac{2v^2\sin \theta \cos \theta }{g}\mathrm{.}$$

Thus, the ratio $\frac{d^{\mathrm{\prime }}}{d}$ is:

Simplify $\sqrt{1.25}$:

$$\sqrt{1.25}=\sqrt{\frac{5}{4}}=\frac{\sqrt{5}}{2}\mathrm{.}$$

Substitute into the ratio:

$$\frac{d^{\mathrm{\prime }}}{d}=\frac{1+\frac{\sqrt{5}}{2}}{2}=\frac{2+\sqrt{5}}{4}\mathrm{.}$$

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Step 9: Value of $n$

The value of $n$ is:

$$n=\frac{2+\sqrt{5}}{4}\mathrm{.}$$

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Final Answer:

The value of $n$ is:

$$n=\frac{2+\sqrt{5}}{4}\mathrm{.}$$