Question

02. A bullet is fired horizontally with a velocity of 200 msIf acceleration due to gravity is 10 ms2? in first second itwill fall through a height of(a) 5'm(b) 10 m20 m(d) 200 m. please explain the answer​

Answer 1

Answer:

Given:

A bullet is fired at a velocity of 200 m/s in the horizontal direction.

Acceleration due to gravity = 10 m/s²

To find:

Height by which the ball will fall in the 1st second.

Concept:

This is a case of height to Ground Projectile . The Projectile (bullet) is projected horizontally with a velocity and due to gravitational acceleration, it moves downwards and reaches the ground.

But during projection , the vertical component of velocity was zero.

Calculation:

Let height by which it falls be h :

$\bigstar \: \: h = u_{y}t + \frac{1}{2} g {t}^{2}$

$= > h = (0 \times 1) + (\frac{1}{2} \times 10 \times {1}^{2} )$

$= > h = 5 \: m$

So final answer is :

$\: \: \: \: \: \: \: \: \: \: \boxed{ \blue{ \huge{ \bold{h = 5 \: m}}}}$

Answer 2

Solution :

✴ Given :-

▪ Intial horizontal velocity of bullet = 200mps

▪ Acceleration due to gravity = 10m/s$^2$

✴ To Find :-

▪ Verticle distance covered by bullet in first second.

✴ Concept :-

✏ This question is completely based on concept of projectile motion (Height to ground).

✏ In this case, verticle velocity of bullet is zero.

✏ Gravitational acceleration acts in downward direction.

✏ We can solve this question with the help of third equation of kinematics.

✴ Calculation :-

$\implies\bf\:H_y=u_yt+\dfrac{1}{2}gt^2\\ \\ \implies\sf\:H_y=(0\times 1)+(\dfrac{1}{2}\times 10\times 1^2)\\ \\ \implies\underline{\underline{\bf{\red{H_y=5m}}}}$