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Grade 11Mechanics

a stone is projected from the ground in such a direction so as to hit a bird sitting on the top of a telegraph post of height h and attains a maximum height of 2h above the ground. if at the instant of projection the bird were to fly away horizontally with the uniform speed. then find the ratio between the horizontal velocity of the bird and the stone if the stone hits the bird while descending

Sparsh Goel

10 Years agoGrade 11

3 Answers

varun

10 Years ago

for heigt of maximum by projectile will be R tan theta =4H and......

R : radius of projectile

H: height of projectile

Piyush Kumar Behera

8 Years ago

Let Vy be the vertical component of the stone's velocity and Vx be it's horizontal velocity.

Max height = 2H = (Vy^2)/2g

Vy^2 = 4gH

Now, the time at which the stone reaches height H. There will be two such times. Let them be T and T', where T > T'.

H = (Vy)t - g(t^2)/2

Solutions of this are T, T'.

T = Vy + sqrt[Vy^2 - 2gH]/g

We need T - T' for the time between the 2 heights.

T - T' = 2sqrt{Vy^2 - 2gH}/g

Finally, (velocity of bird)/(Vx) = (T - T')/T = 2/[(sqrt2) + 1]

Aakarsh

5 Years ago

Time taken to reach at height '2h' from ground is equivalent to time taken to reach on ground from max. height.

So time taken by the stone to reach at height 'h'=

t=√[2.(2h/g)] +_ √(2h/g)

t1=(2-√2)√(h/g)

t2=(2+√2)√(h/g)

Let velocity of bird be v'

Let horizontal component of velocity of stone be v

v'.t2 = v(t2-t1)

v'/v = 2/(√2+1).

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