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

Answer:

Using Newton’s laws of motion, we have

H= (1/2)gt^2. Using H= 1m

1= (1/2) x 9.8 x t^2

Hence, t= square root of 10/49 = 0.45 seconds (approx.)

Now, also velocity with which it hits the ground,

v= square root of 2gh= square root of 2 x 9.8 x 1= 4.4 m/s (approx.)

Explanation:

hope it helps you dear

Answer 2

Answer:

• $t = 0.45\: seconds$

• $v=4.47 \:m/s$

Explanation:

$Given:-$

• Height = 1m

• Initial velocity = 0 m/s

• Gravitational acceleration = 10 m/s²

$ToFind:-$

• Time taken & Final velocity

$Solution :-$

$\underline{By\: using\: kinetic\: energy\: formula}$

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

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

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

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

$\implies \:\: t = 0.45\: seconds$

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

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

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

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

$\therefore$ Time taken is 0.45 seconds and the final velocity is 4.47m/s