Question

A ball dropped from the top of a tower, falls first half of height in 5 s. The total height of tower is(g = 10 m/s2)​

Answer 1

Given :-

▪ A ball is dropped from the top of a tower which is at a height, say h. It falls first half of height i.e., h/2 in time t = 5s

▪ Acceleration due to gravity, g = 10 m/s²

To Find :-

▪ Total height of the tower, or h

Solution :-

We have assumed the height of the tower to be h meters, and it took the ball 5 seconds to reach h/2 . Since it is a free fall, so the initial velocity, u = 0 m/s.

Let us find the height,

Using second second equation of motion, height can be calculated as

⇒ s = ut + 1/2 gt²

⇒ h/2 = 0×5 + 1/2 × 10 × 25

⇒ h/2 = 5 × 25

⇒ h = 10 × 25

⇒ h = 250 m

Hence, The height of the tower is 250 m.

Some Information :-

Other equations of motion,

• 1st : v = u + at
• 2nd : s = ut + 1/2 at²
• 3rd : 2as = v² - u²

Where,

• u = Initial velocity,
• v = final velocity
• a = acceleration
• s = displacement
• t = time

Answer 2

Answer:

Let's draw the figure (I used ms paint).

I tried to draw Burj-khalifa as the tower... I am bad at drawing.

$\bigstar \framebox{GIVEN}$

The ball is dropped from A and it reaches the midpoint C.

• Time = 5 seconds.

• Acceleration due to gravity (g) = 10ms⁻².

• Initial velocity (u) = 0ms⁻¹.

We need to find the height of the tower.

$\bigstar \large \framebox{SOLUTION:}$

Recall the three equations of motion.

$\star \: v=u+at\\\star \: s=ut+\frac{1}{2} at^2\\\star \: 2as = v^2-u^2$

It is apt to use the 1st equation to find out final velocity. Then we can use the third formula to find the distance.

$\to v=u+at$

$\to v=0+10\times 5$

$\to v=50ms^{-1}$

Now, use the third formula:

$\to 2as=v^2-u^2$

$\to 2\times 10\times s=50^2-0^2$

$\to s=2500 \div 20$

$\to s=125 m.$

We just found the length of AC. we know that AC=BC. So, BC=125m.

Total height = BC+AC

= 125+125

=250m.

$\huge \framebox {HOPE THIS HELPS :D}$