Question

A ball is dropped from rest from a height of 40 m above the ground. The height of the ball (from
the ground) after 1 second will be (neglect air friction, g = 10 m/s)
(A) 40 m from the ground
(8) 35 m from the ground
(C) 25 m from the ground
(D) 15 m from the ground​

Answer 1

Answer:

According to the question,

• Initial Velocity (u) = 0 m/s
• Height = 40 m from ground.
• Acceleration (a) = 10 m/s² = g
• Time = 1 second

According to the second equation of motion,

$\rightarrow s = ut + \dfrac{1}{2} at^2$

Now, we are required to find the distance (s) traveled by the ball after 1 second. Substituting the given values in the equation, we get:

$\rightarrow s = 0(1) + \dfrac{1}{2} \times10\times1^2\\\\\rightarrow s = 0 + 5 = 5 \:m$

Therefore in 1 second the ball has traveled 5 m. Therefore the height of the ball from ground is:

→ h = 40 m - 5 m

→ h = 35 m

Therefore the ball is at a height of 35 m from the ground.

Answer 2

Answer:

35 m

Explanation:

Given :

Initial velocity = 0 m / sec

Time = 1 sec

Acceleration due to gravity = 10 m / sec²

Height = 40 m

We have :

h' = u t + 1 / 2 a t²

h' = 0 + 1 / 2 × 10 × 1²

h' = 5 m .

Height of the ball from ground = h - h' = > 40 - 5 = 35 m .

Hence we get answer.