Name: Projectile Motion From a Height
Uploaded: 2019-07-19
Duration: 7 min 6 s
Description: In this video, let us analyse the projectile motion of an object thrown upwards at an angle from a height. Watch this video to learn more about this.

Question

An aeroplane has to go from a point $A$ to another point $B$ , $500 km$ away due $30 °$ east of north. The wind is blowing due north at a speed of $20 m s^{- 1}$ . The air-speed of the plane is $150 m s^{- 1}$ .

(a) Find the direction in which the pilot should head the plane to reach the point $B$ .

(b) Find the time taken by the plane to go from $A$ to $B$ .

Accepted Answer

Given,

Velocity of plane $(v_{p}) = 150 m s^{- 1}$

Velocity of wind $(v_{R}) = 20 m s^{- 1}$

(a) If the aeroplane wants to reach its destination, then its net velocity must be along $A B$ .

In other words, the net velocity perpendicular to line $A B$ must be zero.

Using the diagram and the above statement, we can write,

${(v_{⊥})}_{net} = 0 \Rightarrow v_{R} \sin 30 ° - v_{p} sinθ = 0 \Rightarrow 20 \sin 30 ° - 150 sinθ = 0 \Rightarrow sinθ = \frac{1}{15} \Rightarrow θ = \sin^{- 1} (\frac{1}{15}) = 3 ° 48' .$

To reach $B$ , the plane will fly at an angle of $3 ° 48'$ east of line $A B$ .

(b) If the time taken by the plane to reach from $A$ to $B$ is $t$ , then,

$t = \frac{Displacement between A and B}{The net velocity along A B}$ .

Using the diagram and the angle deduced in the previous part,

$t = \frac{500 \times 1000}{v_{R} \cos 30 ° + v_{p} \cos θ} \Rightarrow t = \frac{500 \times 1000}{20 \cos 30 ° + 150 \cos θ}$ .

As we know, if $\sin θ = \frac{1}{15}$ , then,

$\cos θ = \sqrt{1 - \sin^{2} θ} = \sqrt{1 - {(\frac{1}{15})}^{2}} = \frac{\sqrt{224}}{15}$ .

Substituting the value of $cosθ$ and calculating the value of $t$ , we get,

$t = 2994 s = 2994 \times \frac{1}{60} \min \approx 50 \min$ .

Important Questions on Rest and Motion : Kinematics

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