Question

a ball is dropped from a tower. in last second of its motion it travels a distance of 15m. find the height of the tower

Answer 1

ANSWER ::

Let the total height of the tower = h metre

The ball travelled in last 1 sec=15m

So

The time taken to cover distance of( h-15)=t-1sec

Now

h =1/2gt²

h =1/2gt²it gives h=5t².....(1)

Again

h-15=1/2g(t-1)²

h-15=1/2g(t-1)²5t²-15=5(t-1)² - - - - - - -( 2)

Solving (1) & (2) we get

t=2sec

Putting this value in (1)

h =20m

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Answer 2

Answer

• 20 m

Given

• A ball is dropped from a tower. in last second of its motion it travels a distance of 15 m

To Find

• Height of the tower

Solution

Initial velocity , u = 0 m/s

Distance , s = 15 m

Use 2nd equation of motion .

⇒ s = ut + 1/2 at²

⇒ s = 1/2(10) t²

⇒ t = √(s/5)

At last second , t =  √(s/5) it travels 15 m.

$\implies \sf S_n=u+\dfrac{a}{2}[2n-1]\\\\\implies \sf 15=0+\dfrac{10}{2}[2\bigg(\sqrt{\dfrac{s}{5}}\bigg)-1]\\\\\implies \sf 15=10\sqrt{\dfrac{s}{5}}-5\\\\\implies \sf 20=10\sqrt{\dfrac{s}{5}}\\\\\implies \sf 2=\sqrt{\dfrac{s}{5}}\\\\\implies \sf 4=\dfrac{s}{5}\\\\\implies \sf s=20\ m$

So , the height of the tower is 20 m