Question

An object is dropped from the top of a tower. It travels a distance 'x' in the first
second of its motion and a distance '7x' in the last second. Height of the tower
is : (take g = 10 m/s)
(1) 60 m
(2) 70 m
(3) 80 m
(4) 90 m​

Answer 1

Answer:

Height of tower = 80 m

Explanation:

Given:

• The body travels a distance of x m in the first second
• The body travels a distance of 7x m in the last second.
• Initial velocity of the body = 0 m/s

To Find:

• Height of the tower

Solution:

First we have to find the distance travelled in the first second.

By the second equation of motion we know that,

s = ut + 1/2 × a × t²

where u = initial velocity

a = acceleration = g

t= time taken

Substituting the data we get,

x = 0 × 1 + 1/2 × 10 × 1²

x = 5 m

Hence the distance travelled by the body in the first second is 5 m.

By given,

Distance travelled in the last second = 7x

Distance travelled in the last second = 7 × 5 = 35 m

Also,

If the total time taken to complete the journey is t, the time taken till the last second would be t - 1

Therefore,

Distance travelled in the last second = Total distance travelled - Distance travelled till the last second.

Distance travelled in the last second  = ut + 1/2 at² - u(t - 1) + 1/2 a (t - 1)²

Since u is 0,

Distance travelled in the last second = 1/2 at² - 1/2 a (t - 1)²

Taking 1/2 a as common,

Distance travelled in the last second = 1/2 a (t² - (t - 1)² )

1/2 × 10 (t² - t² + 2t - 1) = 35

5 (2t - 1) = 35

2t - 1 = 7

2t = 8

t = 4

So the total time taken for the journey is 4 seconds.

Now again using the second equation of motion,

s = ut + 1/2 × a × t²

Substitute the value for t,

s = 1/2 × 10 × 4²

s = 16 × 5

s = 80 m

Hence the height of the tower is 80 m.

Therefore option 3 is correct.