Question

A stone falls from a tower and travels 100m in the last second before it reaches the ground .Find the height of the tower

Answer 1

u=0, a=g=9.8m/s^2
S(distance in last second)=u+0.5a(2n-1)
100=0+0.5×9.8(2n-1)
Therefore, n=10.7

That means,total time travel stone took to reach ground=10.7 s

Now, height=ut + 0.5at^2
h=0×t + 0.5 × 9.8 ×(10.7)^2
h=561 m

Answer 2

Concept:

• One-dimensional motion
• Kinematic equations

Given:

• The stone falls, so its initial velocity u = 0 m/s
• The stone travels 100 m in the last second
• The acceleration due to gravity g = 10 m/s^2
• The total time of motion = t

Find:

• The height of the tower

Solution:

We know the kinematic equation

s = ut+1/2 at^2

Height of tower h = ut+1/2gt^2

h = 0 (t) + 1/2 (10) t^2

h = 5t^2

Distance covered from time = 0 to time = t-1

s1 = 0 (t) +1/2 (10) (t-1)^2

s1 = 5 (t-1)^2

s1 = 5 (t^2-2t +1)

s1 = 5t^2 -10t +5

Distance covered in the last second s2 = h-s1 = 100 m

s2 = h - s1

s2 =  5t^2 - (5t^2 -10t +5)

100 = 5t^2 - 5t^2 +10t - 5

100 = 10t -5

105 = 10t

t = 105/10 = 10.5 s

h = 5t^2 = 5 (10.5)^2 = 551.25 m

The height of the tower is 551.25 m.

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