Question

A projectile takes time t to reach from A to B. Distance AB is equal to

Answer 1

Hint: When the object is thrown at some angle from the surface of the Earth. The object covers the curved path and then comes back to Earth. This type of motion is known as the projectile motion.

Complete step-by-step answer:
The projectile is thrown from point A at some angle. It reaches point B after time t.
When the object is projected at a certain angle. The object moves along the x and y axis both. The speed of the projectile in horizontal direction is known as the horizontal component of velocity. The speed of the projectile in vertical direction is known as the vertical component of velocity.

The projectile is making an angle of 600 with the horizontal axis.
The horizontal component of the velocity is given as:
$\begin{aligned} & {{u}_{H}}=u\cos 60 \\\ & {{u}_{H}}=\dfrac{u}{2} \\\ \end{aligned}$

The distance travelled by the projectile horizontally is given as:
Distance = Velocity (time)
$\begin{aligned} & A{B}'={{u}_{H}}t \\\ & A{B}'=\dfrac{u}{2}t \\\ & \\\ \end{aligned}$
Now consider triangle ABB’:
$\begin{aligned} & \cos 30=\dfrac{A{B}'}{AB} \\\ & AB=A{B}'\sec 30 \\\ & AB=\dfrac{ut}{2}\times \dfrac{2}{\sqrt{3}} \\\ & AB=\dfrac{ut}{\sqrt{3}} \\\ \end{aligned}$
Correct option is A.

Note: The horizontal component of the velocity is given as:
${{u}_{H}}=u\cos 60$
The vertical component of the velocity is given as:
${{u}_{V}}=u\sin 60$
The two components of the velocity are different from each other.