Question

For 25 PTS ❤❤...From the top of a structure 100 m high a body is projected vertically upwards with a velocity of 39.2 m/sec .Find velocity with which it strikes the ground.? (Ans is 59.13m/sec ...I need steps please)​

Answer 1

Answer:

• The velocity (v) with it strikes is 59.4 m/s

Given:

1. Height of structure (H) = 100 m.
2. Initial velocity (u) = 39.2 m/s.

Explanation:

$\rule{300}{1.5}$

In this case as ball is thrown vertically upwards from the structure, so first it will reach to its maximum height before it falls down. Lets find the Maximum height achieved by the body.

From the third kinematic equation we know,

$\large\bigstar\;\underline{\boxed{\sf v^{2}-u^{2}=2\;a\;s}}$

Here,

• v Denotes final velocity.
• u Denotes Initial velocity.
• a Denotes acceleration.
• s Denotes displacement/height.

Substituting the values,

(The final velocity will be zero as at highest position the velocity is zero, and acceleration due to gravity will be negative as it is moving against gravity.)

$\longrightarrow\sf \bigg(0\bigg)^{2}-\bigg(39.2\bigg)^{2}=2\times -10\times S\\\\\\\\\longrightarrow\sf 0-1536.64=-20\times S\\\\\\\\\longrightarrow\sf -1536.64=-20\;S\\\\\\\longrightarrow\sf S=\dfrac{1536.64}{20}\\\\\\\longrightarrow\sf S=76.832\\\\\\\longrightarrow\sf S=76.832\;m$

∴ We got the maximum height as 76.832 m.

$\rule{300}{1.5}$

$\rule{300}{1.5}$

Now, the body will fall from this height So, now total height which body has to cover is => 100 + 76.832 = 176.832 m.

From third kinematic equation, we know,

$\large\bigstar\;\underline{\boxed{\sf v^{2}-u^{2}=2\;a\;s}}$

Here,

• v Denotes final velocity.
• u Denotes Initial velocity.
• a Denotes acceleration.
• s Denotes displacement/height.

Substituting the values,

(Here Initial velocity will be zero.)

$\longrightarrow\sf \bigg(v\bigg)^{2}-\bigg(0\bigg)^{2}=2\times 10\times 176.832\\\\\\\\\longrightarrow\sf \bigg(v\bigg)^{2}=20\times 176.832\\\\\\\\\longrightarrow\sf \bigg(v\bigg)^{2}=3536.64\\\\\\\longrightarrow\sf v=\sqrt{3536.64}\\\\\\\longrightarrow\sf v=59.4\\\\\\\longrightarrow\large{\underline{\boxed{\red{\sf v=59.4\;m/s}}}}$

∴ The velocity (v) with it strikes is 59.4 m/s.

$\rule{300}{1.5}$

Answer 2

Answer:

59.4 ms is the correct answer hopes it helps you