Question 54: Motion in 2D & Vectors

Question

The range of the projectile projected at an angle of 15$^\circ$ with horizontal is 50 m. If the projectile is projected with same velocity at an angle of 45$^\circ$ with horizontal, then its range will be

Accepted Answer

The range $R$ of a projectile launched with an initial speed $v$ and at an angle $	heta$ to the horizontal is given by:
$R = \frac{v^2}{g} \sin(2	heta)$
where $g$ is the acceleration due to gravity.
From this equation, we can see that the range is dependent on the sine of twice the launch angle.
Given that the range at $15^\circ$ is $50$ m, if we launch the projectile at $45^\circ$ with the same velocity, we can compare the ranges by comparing $\sin(2 	imes 15^\circ)$ and $\sin(2 	imes 45^\circ)$:
$\sin(30^\circ) = \frac{1}{2}$
$\sin(90^\circ) = 1$
Therefore, the range at $45^\circ$ will be twice the range at $15^\circ$, because $\sin(90^\circ)$ is twice as large as $\sin(30^\circ)$.
So, the range when the projectile is launched at $45^\circ$ will be $2 	imes 50$ m = $100$ m.
So, $100$ m is the correct answer.

Answer

50$\sqrt2$ m

Answer

100 m

Answer

100$\sqrt2$ m

Answer

50 m

Motion in 2D & Vectors

The range of the projectile projected at an angle of 15^\circ with horizontal is 50 m. If the projectile is projected with same velocity at an angle of 45^\circ with horizontal, then its range will be

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