Question

A body is dropped from a height of 320m. The acceleration due to the gravity is
10m/s²?

(a) How long does it take to reach the ground?

(b) What is the velocity with which it will strike the ground?​

Answer 1

Answer:

How long does it take to reach the ground?

→ 8 seconds

What is the velocity with which it will strike the ground?

→ 80 m/s

Explanation:

As per the provided information in the given question, we have :

• Height (h) = 320 m
• Acceleration due to gravity (g) = 10 m/s²

We are asked to calculate time taken by it to reach the ground and the velocity with which it will strike the ground.

Basically, in order to tackle this problem, we need to know some points.

Clarification :

• When the body is thrown vertically downwards (gently),then the initial velocity of the body is 0.[Because it was at rest.]

And, when the body is thrown vertically upwards,then the final velocity it 0.

• When the body is thrown in upward direction, then the acceleration is taken negative because it acts in opposite direction as compared to the direction in which gravity acts.

• When the body is thrown in downward direction, then the acceleration is taken positive.

Now, coming back to the question. Through the help of these points, we can say that :

Here,

• Height (h) = 320 m
• Acceleration due to gravity (g) = +10 m/s²
• Initial velocity (u) = 0

Finding time taken by it to reach the ground :

By using the second equation of motion for frèèly falling bodies,

$\\ \twoheadrightarrow \quad \sf{ s = ut + \dfrac{1}{2}gt^2}$

• s denotes distance
• u denotes initial velocity
• g denotes acceleration due to gravity
• t denotes time

Substituting the values,

$\\ \twoheadrightarrow \quad \sf{ 320 = 0(t) + \dfrac{1}{2}(10) \times t^2}$

$\\ \twoheadrightarrow \quad \sf{ 320 = 5 \times t^2}$

$\\ \twoheadrightarrow \quad \sf{ \dfrac{320 }{5}= 5 \times t^2}$

$\\ \twoheadrightarrow \quad \sf{ 64= t^2}$

$\\ \twoheadrightarrow \quad \sf{ \sqrt{64}= t}$

$\\ \twoheadrightarrow \quad \bf \underline{ 8\; seconds= t}$

Therefore, it takes 8 seconds to reach the ground.

Finding the velocity with which it will strike the ground :

Here, the velocity with which it will strike the ground is actually the final velocity of the body.

By using the first equation of motion for frèèly falling bodies,

$\\ \twoheadrightarrow \quad \sf{ v = u + gt}$

• v denotes final velocity
• u denotes initial velocity
• g denotes acceleration due to gravity
• t denotes time

$\\ \twoheadrightarrow \quad \sf{ v =0 + 10(8)}$

$\\ \twoheadrightarrow \quad \sf{ v =0 + 80}$

$\\ \twoheadrightarrow \quad \bf \underline{ v = 80 \; m/s}$

Therefore, the velocity with which it will strike the ground is 80 m/s.