Question

An Object is released from 20m height then calculate duration of time to reach the ground​

Answer 1

⚘ Question ::

• An Object is released from 20 m height then calculate duration of time to reach the ground.

⚘ Answer ::

• Duration of time to reach the ground is 2 seconds.

Explanation:

⚘ Given that ::

• Distance covered (s) = 20 m
• Initial velocity (u) = 0 m/s
• Acceleration (a) = Acceleration due to gravity = 10 m/s²

⚘ To Find ::

• Time taken by object to reach the ground (t)?

⚘ Solution ::

• Using second equation of motion ::

We know that,

$\large{\boxed{\underline{\underline{\sf{\pink{\bigstar\:\Bigg\{s = ut} \purple{+ \dfrac{1}{2}\:at^2\Bigg\}\:\bigstar}}}}}}$

Where,

• s denotes distance covered
• u denotes initial velocity
• a denotes acceleration
• t denotes time taken

Given,

• s = 20 m
• u = 0 m/s
• a = 10 m/s²

According to the question by using the formula we get,

⇒ $\sf 20 = (0\:\times\:t) + \dfrac{1}{\cancel{2}}\:\times\:\cancel{10}\:\times\:(t)^2$

⇒ $\sf 20 = 0 + 5\:\times\:t^2$

⇒ $\sf 5t^2 = 20$

⇒ $\sf t^2 = {\cancel{\dfrac{20}{5}}}$

⇒ $\sf t^2 = 4$

⇒ $\sf t = \sqrt{4}$

⇒ $\sf t = \sqrt{\underline{2\:\times\:2}}$

➠ $\large{\boxed{\boxed{\bf{\red{t} \blue{=} \pink{2}\:\green{s}}}}}\:\bigstar$

∴ Duration of time to reach the ground is 2 seconds.

⚘ Know More ::

$\clubsuit$ Three equations of motion :

1. v = u + at
2. s = ut + ½ at²ㅤㅤㅤ[Used above]
3. v² = u² + 2as

$\clubsuit$ Some important definitions :

• Acceleration

Acceleration is the process where velocity changes. Since, velocity is the speed and it has some direction. So, change in velocity is considered as acceleration.

• Initial velocity

Initial velocity is the velocity of the object before the effect of acceleration.

• Final velocity

After the effect of acceleration, velocity of the object changes. The new velocity gained by the object is known as final velocity.

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Answer 2

Time taken by object to reach the ground (t)?

⚘ Solution ::

Using second equation of motion ::

We know that,

\large{\boxed{\underline{\underline{\sf{\pink{\bigstar\:\Bigg\{s = ut} \purple{+ \dfrac{1}{2}\:at^2\Bigg\}\:\bigstar}}}}}}★{s=ut+21at2}★

Where,

s denotes distance covered

u denotes initial velocity

a denotes acceleration

t denotes time taken

Given,

s = 20 m

u = 0 m/s

a = 10 m/s²

According to the question by using the formula we get,

⇒ \sf 20 = (0\:\times\:t) + \dfrac{1}{\cancel{2}}\:\times\:\cancel{10}\:\times\:(t)^220=(0×t)+21×10×(t)2

⇒ \sf 20 = 0 + 5\:\times\:t^220=0+5×t2

⇒ \sf 5t^2 = 205t2=20

⇒ \sf t^2 = {\cancel{\dfrac{20}{5}}}t2=520

⇒ \sf t^2 = 4t2=4

⇒ \sf t = \sqrt{4}t=4

⇒ \sf t = \sqrt{\underline{2\:\times\:2}}t=2×2

➠ \large{\boxed{\boxed{\bf{\red{t} \blue{=} \pink{2}\:\green{s}}}}}\:\bigstart=2s★