Question

A ball is dropped from rest. The acceleration due to gravity is 10m s -2 and the time it takes for the ball to reach the ground is 5 seconds. What was the velocity of the ball just before it hit the ground?​

Answer 1

Answer:

50 m/s

Explanation:

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

• Initial velocity of ball (u) = 0 m/s
• Acceleration due to gravity (g) = 10 m/s²
• Time taken (t) = 5 seconds

We are asked to calculate the velocity of the ball just before it hit the ground, i.e its final velocity (v).

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

Substituting the values we have,

$\\ \twoheadrightarrow \quad \sf { v = 0 + 10(5) }$

$\\ \twoheadrightarrow \quad \sf { v = 0 + 50 }$

$\\ \twoheadrightarrow \quad \bf \underline { v = 50 m/s}$

Therefore, the velocity of the ball just before it hit the ground is 50m/s.

More to know!

First equation of motion for frèèly falling bodies :

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

Second equation of motion for frèèly falling bodies :

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

Third equation of motion for frèèly falling bodies :

$\\ \twoheadrightarrow \quad \sf { v^2 -u^2= 2as}$

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