Question
Question: A mango in a tree is located at a horizontal and vertical distance of 30 m and 40 m respectively fro...
A mango in a tree is located at a horizontal and vertical distance of 30 m and 40 m respectively from the point of projection of a stone. Find the minimum speed of the stone so as to hit the mango,
A. u > 30m/s
B. u < 30m/s
C. u = 50m/s
D. u = 40m/s
Solution
The motion of a projectile is a two dimensional motion.
It is separated into two components u cosƟ and u sinƟ.
At a maximum height, initial velocity in downward direction is u sinƟ.
Complete step by step answer:
A stone is being projected to hit a mango in a tree which is located at a horizontal distance of h = 30 m and vertical distance of y = 40 m and let the stone be thrown with a velocity of u at an angle of Ɵ with the horizontal. The velocity u is split into its two components u cosƟ and u sinƟ. The horizontal motion takes place with constant velocity u cosƟ and the distance h covered is in time t.
We can assume that the mango is at its maximum height from the ground and maximum height is 40m.
From kinematic equation, v=usinθ−gt[v=u+at]
0=usinθ−gt [Since the velocity at highest point is zero]
usinθ=gt
t=gusinθ
y=(usinθ)t−21gt2[y=ut+21at2]
y=ut+21gt2
⟹y=(usinθ)(gusinθ)−21g(gusinθ)2
\implies y = \dfrac{{{u^2}{{\sin }^2}\theta }}{g} - \dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}} \\\
⟹y=2gu2sin2θ
u=sin2θ2yg
For minimum speed, the value of sinƟ is maximum and therefore the value of Ɵ is 90⁰. Hence, sin90⁰=1.
u=122×40×10
u=800=28.28ms−1[approx.value]
Thus, minimum velocity required to hit the mango is less than 30ms−1 .
So, the correct answer is “Option b”.
Note:
The initial velocity at the highest point is due to the y component of velocity i.e., v sinƟ.
When the stone reaches the highest point, the final velocity becomes zero as kinetic energy is zero.