Question

A ball is freely from a height if 45m what will be its velocity when it reaches the ground plx help... me​

Answer 1

Given:

A ball is freely dropped from a height of 45m.

We are supposed to find the velocity of the ball just before it hits the ground.

Solution:

Let us write down the 3 basic equations of motion.

1. v = u + at.

2. s = ut + 1/2at².

3. v²-u² = 2as.

We know that, when an object is freely dropped from a height, the initial velocity (u) is 0 and it falls down due to gravity (a=g).

We have the values of u, a and s (=45m).

So, we can use the 3rd equation of motion to find our unknown term(v).

v² - u² = 2as.

v² - (0)² = 2(10)(45) [Assuming g = 10 m/s²].

v² = 900.

v = √900.

v = 30 m/s.

Answer: 30 m/s.

Answer 2

Answer:

Given :-

• A ball is freely from a height of 45 m.

To Find :-

• What is the velocity when it reaches the ground.

Formula Used :-

$\clubsuit$ Third Equation Of Motion :

$\longmapsto \sf\boxed{\bold{\pink{v^2 =\: u^2 + 2gs}}}$

where,

• v = Final Velocity
• u = Initial Velocity
• g = Acceleration due to gravity
• s = Distance Covered

Solution :-

Given :

$\bigstar$ Initial Velocity (u) = 0 m/s

$\bigstar$ Acceleration due to gravity (g) = 10 m/s²

$\bigstar$ Distance Covered (s) = 45 m

According to the question by using the formula we get,

$\longrightarrow \sf v^2 =\: (0)^2 + 2(10)(45)$

$\longrightarrow \sf v^2 =\: 0 + 2 \times 10 \times 45$

$\longrightarrow \sf v^2 - 0 =\: 20 \times 45$

$\longrightarrow \sf v^2 =\: 900$

$\longrightarrow \sf v =\: \sqrt{900}$

$\longrightarrow \sf\bold{\red{v =\: 30\: m/s}}$

$\therefore$ The velocity when it reaches the ground is 30 m/s .