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Question: In a game of angry birds, the bluebird is projected with an angle of \({60^ \circ }\) with a velocit...

In a game of angry birds, the bluebird is projected with an angle of 60{60^ \circ } with a velocity of 6m/s6m/s. After reaching the highest point, the bird splits up into three birds of masses in ratio 2:1:12:1:1. Amongst the three birds, the heaviest bird falls vertically downward with velocity 15m/s15m/s and one bird travels straight. The velocity of the third bird will be:
(A) 9i + 4j\left( A \right){\text{ 9i + 4j}}
(B) 9i + 30.44j\left( B \right){\text{ 9i + 30}}{\text{.44j}}
(C) 3i - 6j\left( C \right){\text{ 3i - 6j}}
(D) 33.44i + 15.22j\left( D \right){\text{ 33}}{\text{.44i + 15}}{\text{.22j}}

Explanation

Solution

Hint
Since we have to find the velocity of the bird. So for this, we have to see the initial momentum and the final momentum. And after equating both the equations we would be able to calculate the velocity of the bird. Also, the angle subtended by the bird with the velocity will be in cosine form.

Complete step by step answer
In the given question the bird is projected with an angle of 60{60^ \circ } and also it is given that the velocity at that time is6m/s6m/s.
So at the highest point, the velocity will beV0cosθ{V_0}\cos \theta
WhereV0{V_0}the velocity at is the highest point and cosθ\cos \theta is the angle subtended between them.
So by putting the values and solving the above solution, we get
6× cosθ\Rightarrow 6 \times {\text{ cos}}\theta
6× 12\Rightarrow 6 \times {\text{ }}\dfrac{1}{2}
3m/s\Rightarrow 3m/s
3i\Rightarrow 3i
Equating the initial momentum and the final momentum, we get
Pi=Pf\Rightarrow {P_i} = {P_f}; Where Pi{P_i}is the Initial momentum and Pf{P_f}is the Final momentum.
So,
(2x+x+x)×3i=(2x)×15.22j+x×3i+x(Ai+Bj)\Rightarrow \left( {2x + x + x} \right) \times 3i = - \left( {2x} \right) \times 15.22j + x \times 3i + x\left( {Ai + Bj} \right)
On solving the above equation, we get
4×3i=2×15.22j+3i+(Ai+Bj)\Rightarrow 4 \times 3i = - 2 \times 15.22j + 3i + \left( {Ai + Bj} \right)
On further solving
9i+30.44j=(Ai+Bj)\Rightarrow 9i + 30.44j = \left( {Ai + Bj} \right)
9i+30.44j9i + 30.44j Is the required velocity and Hence Option (B) will be the correct answer.

Note
The parabolic flight of a projectile, projectile motion on inclined planes. An object that's on the wing once being thrown is called a projectile. As an example, if you throw a ball up in the air at an angle apart from zero degree with the vertical, it follows a recursive path that is named its trajectory. It becomes easier to investigate the trail of a projectile if we tend to create the subsequent two assumptions:
The free-fall acceleration g is constant over the vary of motion and is directed downward, and The impact of air resistance is negligible. With these assumptions, we discover that the flight of a projectile is often a parabola that may be a version of a plot of a polynomial like this.