Question 7: Motion in 2D & Vectors

Question

A projectile is thrown upward at an angle $60^{\circ}$ with the horizontal. The speed of the projectile is $20 \mathrm{~m} / \mathrm{s}$ when its direction of motion is $45^{\circ}$ with the horizontal. The initial speed of the projectile is $\_\_\_\_$ $\mathrm{m} / \mathrm{s}$.

Accepted Answer

When a projectile is launched into the air, it moves in two dimensions: horizontally ( $\mathrm{x}-$ axis ) and vertically ( y - axis).
Along vertical direction gravity constantly pulls the projectile downward. Therefore, there is a constant downward acceleration $\left(\mathrm{a}_{\mathrm{y}}=-\mathrm{g}ight)$.
Along horizontal direction there are no horizontal forces (assuming no air resistance). According to Newton's First Law, an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force. Therefore, the horizontal acceleration is zero ( $\mathrm{a}_{\mathrm{x}}=0$ ).
Because $\mathrm{a}_{\mathrm{x}}=0$, the horizontal component of the projectile's velocity remains perfectly constant throughout its entire journey.
If a projectile is launched with an initial speed u at an angle $	heta$, its horizontal component is
$$ \mathrm{u}_{\mathrm{x}}=\mathrm{u} \cos 	heta $$
At any later point in time, if the projectile has a new speed v and is moving at a new angle $\alpha$, its horizontal component is :
$$ \mathrm{v}_{\mathrm{x}}=\mathrm{v} \cos \alpha $$
Since the horizontal velocity never changes, these two expressions must be equal :
$$ \mathrm{u}_{\mathrm{x}}=\mathrm{v}_{\mathrm{x}} $$
$$\Rightarrow $$ $$ \mathrm{u} \cos 	heta=\mathrm{v} \cos \alpha  $$
Substituting the given values :
$$ \mathrm{u} \cos \left(60^{\circ}ight)=20 \cos \left(45^{\circ}ight) $$
$$\Rightarrow $$ $u\left(\frac{1}{2}ight)=20 	imes\left(\frac{1}{\sqrt{2}}ight)$
$$\Rightarrow $$ $u=20 	imes \frac{2}{\sqrt{2}}$
Therefore, the initial speed of projectile is $20 \sqrt{2} \mathrm{~m} / \mathrm{s}$.
Hence, the correct option is (B).

Answer

$20 \sqrt{3}$

Answer

$20 \sqrt{2}$

Answer

40

Answer

$40 \sqrt{2}$

Motion in 2D & Vectors

A projectile is thrown upward at an angle 60^{\circ} with the horizontal. The speed of the projectile is 20 \mathrm{~m} / \mathrm{s} when its direction of motion is 45^{\circ} with the horizontal. The initial speed of the projectile is \_\_\_\_ \mathrm{m} / \mathrm{s}.

Choose the correct answer: