Question

A stone dropped from the top of a tower, takes 4 s to reach ground level. Calculate i) the final velocity of the stone ii) height of the tower. (take g = 10m/s2)

Answer 1

Explanation:

v=u+at

u=0(due to body is in a rest)

a=10m/s

t=4s

0+10*4=40m/s

a) =40m/s

v2-u2=2as

(40)2-(0)2=2*10*s

1600=20s

s=80m

Answer 2

• The final velocity of the stone = 40 m/s

• The height of the tower = 80 m

Given :

A stone dropped from the top of a tower, takes 4 s to reach ground level

To find :

• The final velocity of the stone
• The height of the tower

Solution :

Step 1 of 3 :

Write down initial velocity , time and acceleration

Here it is given that a stone dropped from the top of a tower, takes 4 s to reach ground level

Initial velocity = u = 0

Time = t = 4s

Acceleration = g = 10 m/s²

Step 2 of 3 :

Calculate final velocity of the stone

Let final velocity of the stone = v m/s

By the given condition

$\displaystyle \sf{ v = u + gt}$

$\displaystyle \sf{ \implies v = 0 + (10 \times 4)}$

$\displaystyle \sf{ \implies v =40 }$

∴ The final velocity of the stone = 40 m/s

Step 3 of 3 :

Calculate height of the tower

Let height of the tower = h meter

Then by 2nd equation of motion we get ,

$\displaystyle \sf{ h = ut + \frac{1}{2}g {t}^{2} }$

$\displaystyle \sf{ \implies h = (0 \times 4) + \frac{1}{2} \times 10 \times {4}^{2} }$

$\displaystyle \sf{ \implies h = 0 + 5 \times 16}$

$\displaystyle \sf{ \implies h = 80 }$

∴ The height of the tower = 80 m

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