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
Answers
Answer:
the answer is option b 5root2
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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