A projectile is fired from horizontal ground with speed $v$ and projection angle $\theta$. 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 $g^{\prime }=\frac{g}{0.81}$, then the new range is $d^{\prime }=nd$. The value of $n$ is _____.

The correct answer is n = 0.95.

We solve this by splitting the trajectory into two parts: the first half (from launch to the highest point, under gravity $g$), and the second half (from the highest point to landing, under the new gravity $g^{\prime }=\frac{g}{0.81}$).

Step 1: Standard range under gravity g

The standard range of a projectile on flat ground is:

$d=\frac{v^{2}sin2\theta }{g}=\frac{2v^2\sin \theta \cos \theta }{g}$

By the symmetry of projectile motion under uniform gravity, the horizontal distance covered during the ascent (launch to highest point) equals exactly half the total range:

$x_1=\frac{d}{2}$

Step 2: Velocity and height at the highest point

At the highest point of the trajectory:

- The vertical velocity component is zero.

- The horizontal velocity component is $v_x=v\cos \theta$ (unchanged throughout).

The maximum height reached is:

$H=\frac{v^2{\sin }^2\theta }{2g}$

Step 3: Time to fall in the new gravity region

Once the projectile enters the new region at the highest point, it must fall a vertical distance $H$ under the new effective gravity $g^{\prime }=\frac{g}{0.81}$. Using $H=\frac{1}{2}{g}^{\prime }{t}_2^2$, the time of descent is:

$t_2=\sqrt{\frac{2H}{g^{\prime }}}=\sqrt{\frac{v^2{\sin }^2\theta }{g}}$

Simplifying:

$t_2=\frac{v\sin \theta }{g}\cdot \sqrt{0.81}=\frac{0.9\cdot v\sin \theta }{g}$

Note that $\sqrt{0.81}=0.9$.

Step 4: Horizontal distance covered in the new region

The horizontal velocity is still $v\cos \theta$ (gravity only acts vertically), so the horizontal distance covered during descent is:

$x_2=v\cos \theta \cdot t_2=v\cos \theta \cdot \frac{0.9\cdot v\sin \theta }{g}=\frac{0.9\cdot v^2\sin \theta \cos \theta }{g}$

Recognizing that $\frac{v^2\sin \theta \cos \theta }{g}=\frac{d}{2}$, we get:

$x_2=0.9\cdot \frac{d}{2}$

Step 5: Calculate the new total range d'

The total new range is the sum of the two horizontal segments:

$d^{\prime }=x_1+x_2=\frac{d}{2}+0.9\cdot \frac{d}{2}=\frac{d}{2}(1+0.9)=\frac{1.9}{2}d=0.95d$

Conclusion:

Since $d^{\prime }=0.95d=nd$, we get:

$n=0.95$

The key insight is that increasing gravity (since $\frac{g}{0.81}>g$) causes the projectile to fall faster in the second half of its journey, reducing the time of flight and thus shortening the horizontal distance covered in that half. The first half of the range is unaffected (it occurred under the original gravity), so the new total range is slightly less than the original, giving n = 0.95.