Question

A STONE IS DROPPED FROM THE ROOF OF A TOWER OF HEIGHT H. THE TOTAL DISTANCE COVERED BY THE STONE IN THE LAST 2 SECONDS OF ITS MOTION IS EQUAL TO THE DISTANCE COVERED BY IT IN THE FIRST FOUR SECONDS. FIND THE HEIGHT OF THE TOWER

Answer 1

Given that,

A ball is dropped from the roof of a tower height h.

Total distance covered it in the last seconds of its motion.

We need to calculate the distance covered in first 4 sec

Using equation of motion

$s=ut+\dfrac{1}{2}gt^2$

Where, u = initial velocity

t = time

g = acceleration due to gravity

Put the value into the formula

$s=0+\dfrac{1}{2}\times9.8\times(4)^2$

$s=78.4\ m$

If the ball takes in second to fall to ground.

The distance covered in nth second.

We need to calculate the value of n

Using equation of motion

$s_{n}=u+\dfrac{g}{2}(2n-1)$

Put the value into the formula

$78.4=0+\dfrac{9.8}{2}(2\times n-1)$

$78.4=9.8n-4.9$

$9.8n=78.4+4.9$

$n=\dfrac{78.4+4.9}{9.8}$

$n=8.5$

We need to calculate the total height covered by the stone

Using equation of motion

$h=ut+\dfrac{1}{2}gt^2$

Put the value into the formula

$h=0+\dfrac{1}{2}\times9.8\times(8.5)^2$

$h=354.0\ m$

Hence, The total height covered by the stone is 354 m

Answer 2

Explanation:

tower height h.

Total distance covered it in the last seconds of its motion.

We need to calculate the distance covered in first 4 sec

Using equation of motion

s=ut+\dfrac{1}{2}gt^2s=ut+

2

1

gt

2

Where, u = initial velocity

t = time

g = acceleration due to gravity

Put the value into the formula

s=0+\dfrac{1}{2}\times9.8\times(4)^2s=0+

2

1

×9.8×(4)

2

s=78.4\ ms=78.4 m

If the ball takes in second to fall to ground.

The distance covered in nth second.

We need to calculate the value of n

Using equation of motion

s_{n}=u+\dfrac{g}{2}(2n-1)s

n

=u+

2

g

(2n−1)

Put the value into the formula

78.4=0+\dfrac{9.8}{2}(2\times n-1)78.4=0+

2

9.8

(2×n−1)

78.4=9.8n-4.978.4=9.8n−4.9

9.8n=78.4+4.99.8n=78.4+4.9

n=\dfrac{78.4+4.9}{9.8}n=

9.8

78.4+4.9