Question

A ball is dropped from the roof of a tower of height 180 m. What is the distance covered by the ball in the
last second? (take g = 10 m/s2)
​

Answer 1

Answer:

Answer is 6

Explanation:

h = 180m

g= 10m/s^2

u =0m/s

According to question;

h = ut +1/2gt^2

180=(0)t+1/2×10×t^2

180=1/2×10×t^2

180×2/10=t^2

36= t^2

t^2=36

t= 6^2(1/2)

t=6sec

Answer 2

The answer is 55 meters

GIVEN

A ball is dropped from the roof of a tower of height 180 m

TO FIND

Distance covered in the last second.

SOLUTION

We can simply solve the above problem as follows;

Height of the tower = 180m

Initial velocity of the ball = 10 m/s²

Acceleration due to gravity = 10m/s

Let the time taken to reach the ground = t

Applying second equation of motion

s = ut + at²/2

180 = 0 + 5t²

t² = 180/5 = 36

t = √36 = 6 seconds

Applying the formula ;

$Sₙ = u + a(n - \frac{1}{2} )$

Where,

n = time

Therefore,

$Sₙ = 0 + 10(6 - \frac{1}{2} )$

= 10( 12 -1/2)

= 10 × 11/2

= 55 meters

Hence, The answer is 55 meters

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