Question

A ball is dropped from a certain height such that the distance travelled in last second is equal to the distance travelled in first three second. find its height

Answer 1

Let us consider that a ball is dropped from height h such that the distance travelled in last second is equal to distance travelled in first three second.

distance travelled in first three second, s = ut + 1/2 at²

here , u = 0, t = 3sec and a = 10m/s² [ acceleration due to gravity]

so, s = 1/2 × 10 × (3)²

= 45 m .....(1)

now find total time taken by ball to reach the ground .

h = 0 + 1/2 gT² [ from formula, s = ut + 1/2 at² ]

T = √{2h/g} .......(2)

so, last second is Tth second.

now, distance travelled by ball in during Tth second, s' = u + a/2(2n - 1) formula,

here, u = 0, a = 10 m/s² and n = T

then, s' = 0 + 10/2(2T - 1)

or, s' = 5(2T - 1) .....(3)

a/c to question, s = s'

so, 45 = 5(2T - 1) [ from equations (1) and (3) ]

or, T = 5

from equation (2),

√{2h/g} = 5

or, 2h/g = 25

or, h = 25g/2 = 125m

hence, height of ball from the ground is 125m

Answer 2

Answer:

125 metres is the height.

Explanation:

The distance covered in first three seconds is :

[ s = 1/2 gt² ]

where 'u' is equal to zero.

where, u = 0

s = 1/2 gt²

s = 1/2 × 10 × 3²

s = 5 × 3²

s = 5 × 9

s = 45 metres

If the ball takes 'n' second to fall to the ground. The distance covered in 'nth' second.

Sn = u + g / 2 × (2n - 1)

Sn = 0 + 10 / 2 × (2n - 1)

Therefore,

45 = 10n - 5

=> 45 + 5 = 10n

=> 50 = 10n

=> n = 50 / 10

.°. n = 5 seconds

Therefore,

h = 1/2 gt²

h = 1/ 2 × 10 × 5²

h = 1 / 2 × 10 × 25

h = 5 × 25

h = 125 metres.

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