Question

A particle thrown up vertically reaches at half of the maximum height in (✓2 – 1) s. The speed of projection of the particle is (g = 10 m/s2


1) 10
2) 5✓2
3) 10✓2​

Answer 1

Answer:

the answer is option b 5root2

Answer 2

Concept:

Throwing a particle upward will take u/g time to stop at its maximum height. Where u is the initial velocity and g is acceleration due to gravity.

Given:

Time taken to reach half of maximum height is (✓2 – 1) seconds.

Value of g = 10 m/s²

Find:

Initial Speed of projection (u)

Solution:

Lets assume (✓2 – 1) as t'

Applying second equation of motion for maximum height

s = ut + (1/2)gt² ---1

Applying second equation of motion for half of maximum height

s/2 = ut' + (1/2)gt'²

Multiplying whole equation with 2

s = 2ut' + 2(1/2)gt'² ---2

Subtracting equation 1 from 2

0 = 2ut' + 2(1/2)gt'² - ( (1/2)gt²  + ut )

0 = 2ut' + gt'² -  (1/2)gt²  - ut

As we know Throwing a particle upward will take u/g time to stop at its maximum height.

0 = 2ut' + gt'² -  (1/2)g(u/g)²  - u(u/g)

0 = 2ut' + gt'² -  (1/2)(u²/g)  - (u²/g)

2ut' + gt'² - (u²/g) ( (1/2) +1 ) = 0

2ut' + gt'² - (3/(2g)) u² = 0

- (3/20) u² + 2(✓2 – 1)u + 10×(✓2 – 1)  = 0

- 0.15 u² + 0.82 u + 4.1  = 0

15 u² - 82 u - 410  = 0

There will be two value of u

u = 8.63 m/s and u = -3.16 m/s

We will reject negative value as initial velocity was in positive direction.

Hence, The speed of the projection of the particle will be 8.63 m/s

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