Question 12: Motion in 2D & Vectors

Question

The two projectiles are projected with the same initial velocities at the $15^{\circ}$ and $30^{\circ}$ with respect to the horizontal. The ratio of their ranges is $1: x$. The value of $x$ is

Accepted Answer

Let u be the initial velocity and $	heta=45^{\circ}$ be the angle of projection.
The initial vertical velocity is $\mathrm{u}_{\mathrm{y}}=\mathrm{u} \sin 	heta$
The ball attains maximum height in $\mathrm{t}_{\max }=2 \mathrm{~s}$.
At the maximum height, the vertical component of the velocity becomes zero $\left(\mathrm{v}_{\mathrm{y}}=0ight)$.
Using the first equation of motion $(\mathrm{v}=\mathrm{u}+\mathrm{at})$ :
$$ \mathrm{v}_{\mathrm{y}}=\mathrm{u}_{\mathrm{y}}-\mathrm{gt} $$
$$\Rightarrow $$ $$ 0=\mathrm{u}_{\mathrm{y}}-(10)(2) \Rightarrow \mathrm{u}_{\mathrm{y}}=20 \mathrm{~m} / \mathrm{s} $$
So, the initial vertical velocity is $20 \mathrm{~m} / \mathrm{s}$.
The ball lands on the roof of the building at $\mathrm{t}=3 \mathrm{~s}$.
This means at $t=3$, the vertical displacement $(y)$ of the ball is equal to the height of the building $(H)$.
Using the second equation of motion $\left(s=u t+\frac{1}{2} a t^2ight)$ :
$$ \mathrm{y}=\mathrm{u}_{\mathrm{y}} \mathrm{t}-\frac{1}{2} \mathrm{gt}^2 $$
Substituting the given values :
1. $y=H$ (Height of the building)
2. $\mathrm{u}_{\mathrm{y}}=20 \mathrm{~m} / \mathrm{s}$ (Calculated in Step 2)
3. $\mathrm{t}=3 \mathrm{~s}$ (Given time of landing)
$$ H=(20)(3)-\frac{1}{2}(10)(3)^2=15 \mathrm{~m} $$
Therefore, the height of the building is 15 m. Hence, the correct option is (D).

Answer

$\sqrt{2}$

Answer

$$ \sqrt{3} $$

Answer

$2 \sqrt{3}$

Answer

$$  \frac{1}{\sqrt{2}} $$

Motion in 2D & Vectors

The two projectiles are projected with the same initial velocities at the 15^{\circ} and 30^{\circ} with respect to the horizontal. The ratio of their ranges is 1: x. The value of x is

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