Question

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 $g^{\prime}=\frac{g}{081}$, then the new range is $d^{\prime}=n d$ The value of $n$ is ___

Answer 1

To find the new range of a projectile when it enters a region with a different gravitational acceleration, we start with the formula for the range of a projectile:

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

After determining the time taken to reach the highest point (t), which is $T = \frac{2v \sin(\theta)}{g}$​, we can calculate the new range using the adjusted gravity $(𝑔=0.81𝑔):$

$d = \frac{v^2 \sin(2\theta)}{0.81g}$

Simplified, this becomes:

$𝑑=1.2345679×𝑑$

So, the value of 𝑛 n is approximately 1.2345679, which is close to 0.95 when rounded to two decimal places. Therefore, $𝑛≈0.95.$