Question

A ball is released from the top of the tower of height 160 m. If it takes T seconds to reach the ground, then the position of ball at T/4 second is
1. 150 m from the ground
2. 40 m from the ground
3. 130 m from the ground
4. 10 m from the ground​

Answer 1

Answer:

Answer will be (1)150m from the ground

Dam correct !!

Explanation:

Here u=0,s=160m,a=g

So by 2nd equation of motion

S=ut+1/2at^2

we have,

160=0+1/2gt^2

--> 160=1/2gt^2      (1)

Now we are asked for t/4 second so

S=ut+1/2at^2    (here u=0,a=g ,t=t/4)

S=0+1/2g(t/4)^2

S=1/2gt^2/16

here we can put (1) that is 1/2gt^2 = 160

S=160/16

S=10m

And now

160-10=150m Hope u understood !!!!

Answer 2

Concept:

• Free fall motion
• One-dimensional motion
• Kinematics equations

Given:

• Height of the tower s = 160 m
• The initial velocity of the ball = 0 m/s
• The time taken for the ball to reach the ground = T
• Acceleration due to gravity g = 10m/s^2

Find:

• The position of the ball at T/4 seconds

Solution:

We know the kinematics equation

s = ut+1/2at^2

160 = 0 +1/2 (10) T^2

160/5 = T^2

T = √32 s = 4√2 s

At t = T/4 = 4√2/4 = √2 s

s = ut+1/2at^2

s = 0 + 1/2 (10) (2)

s = 10m

The ball has covered 10 m from the top of the tower in T/4s

The position of the ball is 160-10 = 150m from the ground

The position of the ball is 150 m from the ground.

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