Question

A ball is dropped from a height of 1 metre .how long does it take to reach the ground and with what speed will it hit the ground g =10 m per second square​

Answer 1

Given :-

• Height (h) = 1m
• Initial velocity (u) = 0 m/s
• Acceleration due to gravity (g) = 10 m/s²

To Find :-

• Time taken (t)
• Final velocity (v)

Solution :-

We know :-

$\boxed{\bf s = ut+ \dfrac{1}{2}gt^2}$

$: \implies \sf 1= 0\times t + \dfrac{1}{2} \times 10 \times t^2$

$: \implies \sf 1= 5t^2$

$: \implies \sf t^2 = 0.2$

$: \implies \bf t = 0.45 s$

We know :-

$\boxed{\bf v^2 = u^2+2gh}$

$: \implies \sf v^2= 0^2+ 2\times 10 \times 1$

$: \implies \sf v^2 = 20$

$: \implies \sf \bf v=4.47 m/s$

Answer 2

Answer:

Given :-

• Height = 1 m
• Initial velocity = 0 m/s
• Acceleration due to gravity = 10 m/s²

To Find :-

• Time taken
• Final Velocity

Solution :-

At first see some keywords

• There are two types of velocity
• 1) Initial velocity
• 2) Final Velocity
• Initial velocity is the continuous velocity of an object
• Final Velocity is the velocity that comes after rest.
• The acceleration due to gravity is 10 m/s²
• SI unit of Acceleration = m/s²

Now,

We know that

s = ut + ½gt²

• S = Distance travelled
• U = initial velocity
• g = Acceleration due to gravity
• t = time

Here,

Height = Distance travelled

1 = 0 × t + ½ × 10 × t²

1 = 0t + 5 × t²

1 = 0t + 5t²

1 = 5t²

1/5 = t²

0.2 = t²

0.44 = t

Now,

Calculating final velocity

v² - u² = 2gs

v² - 0² = 2(10)(1)

v² = 20

v = √20

v = 4.47 m/s