A stone is projected from the ground in such a direction so as to hit ...
Problem:
A stone is projected from the ground in such a direction so as to hit a bird on the top of a telegraph post of height h and attains the maximum height of 3/2h above the ground. If at the instant of projection, the bird were to fly away horizontally with a uniform speed, find the ratio between the horizontal velocity of bird and the horizontal component of velocity of stone, if the stone hits the bird while descending?
Solution:
Initial Parameters:
Let the initial velocity of the stone be v and the angle of projection be θ. The horizontal component of the velocity of the stone will be vcosθ. The time taken for the stone to reach the maximum height will be t = (v sinθ)/g, where g is the acceleration due to gravity.
Maximum Height:
The maximum height attained by the stone is 3/2h. Thus, using the kinematic equation, s = ut + (1/2)at², where s = 3/2h, u = v sinθ, a = -g and t = (v sinθ)/g, we get:
3/2h = (v sinθ)((v sinθ)/g) - (1/2)g((v sinθ)/g)²
Simplifying this equation, we get:
(v² sin²θ)/(2g) = 3/2h
v² sin²θ = 3gh
Horizontal Distance:
The horizontal distance travelled by the stone before hitting the bird will be d = vcosθ(t + t'), where t' is the time taken by the stone to descend from the maximum height to the height of the bird, which is h.
Using the kinematic equation, v² = u² + 2as, where u = v sinθ, a = -g and s = h - (3/2)h = -(1/2)h, we get:
v² sin²θ = 2gh/3
Thus, the horizontal distance travelled by the stone before hitting the bird will be:
d = vcosθ((v sinθ)/g + √((2h)/(3g)))
Ratio of Velocities:
The bird is flying away horizontally with a uniform speed. Let the horizontal velocity of the bird be u. The time taken by the stone to hit the bird will be t + t'.
Thus, using the equation d = ut, we get:
u = d/(t + t') = vcosθ((v sinθ)/g + √((2h)/(3g)))/((v sinθ)/g + 2√((2h)/(3g)))
Thus, the ratio between the horizontal velocity of bird and the horizontal component of velocity of stone will be:
u/(vcosθ) = ((v sinθ)/g + √((2h)/(3g)))/((v sinθ)/g + 2√((2h)/(3g)))
Answer:
The ratio between the horizontal velocity of bird and the horizontal component of velocity of stone will be ((v sinθ)/g + √((2h)/(3g)))/((v sinθ)/g + 2