A projectile is fired from horizontal ground with speed v and projection angle theta. When the accel

Q: Why doesn't the horizontal velocity change in the new region?

Gravity, whether it is $g$ or $g'$, only acts in the vertical direction. Since there is no horizontal acceleration (assuming no air resistance), the horizontal velocity remains constant throughout the entire flight.

Q: Why did we use $H_{max}$ for the second part of the motion?

The problem explicitly states that the gravity changes exactly at the highest point. At this specific moment, the projectile is at its maximum height $H_{max}$ from the ground, and it begins its descent from this exact altitude.

Q: How did $0.81$ come out of the square root as $0.9$?

The term $0.81$ is located in the denominator of the denominator, so algebraically, it flips to the numerator. The square root of $0.81$ is exactly $0.9$, which makes the calculation very clean and intentional by the examiner!

Q: Can we just use the standard range formula for the whole trajectory and plug in $g'$?

No, the standard range formula $R = \frac{v^2 \sin 2\theta}{g}$ assumes a constant acceleration $g$ for the entire duration of the flight. Since $g$ changes halfway through, the symmetry is broken, and we must analyze the two halves separately.

Q: What if the gravity changed before it reached the highest point?

That would be a much more complex problem! You would have to calculate the exact height and the non-zero vertical velocity at that specific moment, and then use those as the initial conditions for the new region.

Question

A projectile is fired from horizontal ground with speed $v$ and projection angle $	heta$. 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 $d' = nd$. The value of $n$ is_______.

Sigma Percentile · Accepted Answer

### The Setup: A Tale of Two Gravities

Imagine you are watching a projectile launch. Under normal circumstances, it traces a perfectly symmetric parabolic path. We know the standard formulas by heart: the total horizontal range is given by $d = \frac{v^2 \sin 2	heta}{g}$, and the maximum height it reaches is $H_{max} = \frac{v^2 \sin^2 	heta}{2g}$.

Because the standard parabola is symmetric, the projectile reaches its peak exactly halfway through its journey. This means the horizontal distance covered up to the highest point is exactly $\frac{d}{2}$.

But this problem introduces a fascinating twist. The universe changes its rules right when the projectile is at its most vulnerable point—the peak of its trajectory.

### The Peak: A Moment of Transition

Let's freeze time at the exact moment the projectile reaches $H_{max}$. What is its state?

At the highest point, the projectile has momentarily stopped moving upwards. Its vertical velocity is exactly zero ($v_y = 0$). However, it hasn't stopped moving entirely. It still possesses its initial horizontal velocity, $v_x = v \cos	heta$. Because gravity only pulls downwards, this horizontal velocity has remained completely untouched and constant since the moment of launch.

Suddenly, as it crosses this peak, it enters a new region. The downward pull of gravity intensifies! The effective acceleration due to gravity is now $g' = \frac{g}{0.81}$.

### The Descent: Falling Faster

Now, we must treat the second half of the journey as a completely new physics problem. We have an object dropped from a height $H_{max}$ with zero initial vertical velocity, but it is now accelerating downwards at $g'$.

How long will it take to hit the ground? We can use the second equation of motion for the vertical drop: $s = ut + \frac{1}{2}at^2$. Since the initial vertical velocity is zero, this simplifies to $H_{max} = \frac{1}{2}g'(t')^2$.

Solving for the new time of descent, $t'$, we get:

$$t' = \sqrt{\frac{2H_{max}}{g'}}$$

Let's substitute our known values into this equation:

$$t' = \sqrt{\frac{2 \left( \frac{v^2 \sin^2 	heta}{2g} ight)}{\frac{g}{0.81}}}$$

The $2$ in the numerator and denominator beautifully cancel out. The $0.81$ in the denominator of the denominator flips up to the top.

$$t' = \sqrt{\frac{0.81 v^2 \sin^2 	heta}{g^2}}$$

Taking the square root yields a remarkably clean result:

$$t' = \frac{0.9 v \sin 	heta}{g}$$

### The Final Stretch: Adding It All Up

While the projectile is plummeting towards the ground faster than it normally would, it is still moving horizontally at its constant speed of $v \cos	heta$.

The horizontal distance it covers during this new descent, let's call it $d_1$, is simply the product of its horizontal speed and the new time of descent:

$$d_1 = (v \cos 	heta) 	imes t'$$

$$d_1 = (v \cos 	heta) \left( \frac{0.9 v \sin 	heta}{g} ight)$$

Rearranging the terms, we get:

$$d_1 = 0.9 \left( \frac{v^2 \sin 	heta \cos 	heta}{g} ight)$$

We can recognize the structure of the double angle formula here. Since $\sin 2	heta = 2 \sin 	heta \cos 	heta$, we can rewrite the expression as:

$$d_1 = 0.45 \left( \frac{v^2 \sin 2	heta}{g} ight)$$

Notice that the term in the parentheses is exactly our original range, $d$! Therefore, $d_1 = 0.45d$.

Finally, to find the total new range $d'$, we add the horizontal distance covered before the peak to the horizontal distance covered after the peak:

$$d' = \frac{d}{2} + d_1$$

$$d' = 0.5d + 0.45d = 0.95d$$

The problem states that the new range is $nd$. By comparing our result, we can confidently conclude that **$n = 0.95$**.

Enter Numerical Value:

Visualized Solution

$d = \frac{v ^{2} s i n 2 θ}{g}$

$v_{y} = 0, v_{x} = v cos θ$

$g^{'} = \frac{g}{0.81}$

$t^{'} = \frac{2 H _{ma x}}{g ^{'}}$

$d_{1} = u_{x} t^{'}$

$d^{'} = \frac{d}{2} + d_{1}$

$g^{'} < g ⟹ d^{'} > d$

The Sigma Insight: Projectile Motion

The Setup

The Peak

The Descent

The Final Stretch

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } d=gv2sin2θ​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } vy​=0,vx​=vcosθ

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g′=0.81g​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t′=g′2Hmax​​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } d1​=ux​t′

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } d′=2d​+d1​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g′<g⟹d′>d