Question 142: Motion in 2D & Vectors

Question

The maximum height reached by a projectile is $$64 \mathrm{~m}$$. If the initial velocity is halved, the new maximum height of the projectile is ______ $$\mathrm{m}$$.

Accepted Answer

To solve this problem, we first need to understand the formula that relates the maximum height $H$ reached by a projectile to its initial velocity $v_0$ and the acceleration due to gravity $g$:
$H = \frac{v_0^2 \sin^2(	heta)}{2g}$
where:
$H$ is the maximum height,
$v_0$ is the initial velocity of the projectile,
$	heta$ is the angle of projection with the horizontal, and
$g$ is the acceleration due to gravity, which is approximately $9.8 \, \mathrm{m/s^2}$.
Given that the maximum height attained by the projectile is $64 \, \mathrm{m}$, we can write:
$64 = \frac{v_0^2 \sin^2(	heta)}{2 \cdot 9.8}$
Now, we are asked to find the new maximum height if the initial velocity is halved. Let's denote the new initial velocity as $v'_0 = \frac{v_0}{2}$. Using the formula for maximum height again, we get:
$H' = \frac{{v'_0}^2 \sin^2(	heta)}{2g}$
Substituting $v'_0 = \frac{v_0}{2}$ into this equation:
$H' = \frac{(\frac{v_0}{2})^2 \sin^2(	heta)}{2g} = \frac{v_0^2 \sin^2(	heta)}{2 \cdot 4g} = \frac{1}{4} \cdot \frac{v_0^2 \sin^2(	heta)}{2g}$
Since we know the original height:
$64 = \frac{v_0^2 \sin^2(	heta)}{2 \cdot 9.8}$
Substituting this value into our equation for $H'$, we find:
$H' = \frac{1}{4} \cdot 64 = 16 \, \mathrm{m}$
Therefore, if the initial velocity of the projectile is halved, the new maximum height reached by the projectile would be $16 \, \mathrm{m}$.

Motion in 2D & Vectors

The maximum height reached by a projectile is $64 \mathrm{~m}. If the initial velocity is halved, the new maximum height of the projectile is ______ \mathrm{m}$.

Choose the correct answer: