Projectile Motion (11Acd03)

Sheet 05

  1. What will be the angle of projection for a projectile, whose range \(\left(R\right)\) is \(n\) times the maximum height \(\left(H\right)\)?

    Ans: \(\theta=tan^{.-1}\left(\frac{4}{n}\right)\).

  2. A projectile of mass \(m\) is thrown with a velocity \(v\) from the ground at angle \(45^{o}\) with the horizontal. What is the change in momentum between leaving and arriving back at the ground?

    Ans: \(\sqrt{2}mv\).

  3. If, a projectile has a constant initial speed and angle of projection, then what is the relation between the change in the horizontal range due to the change in gravity?

    Ans: \(\frac{dR}{R}=\frac{-dg}{g}\).

  4. A helicopter on a flood relief mission flying horizontally with a speed \(u\) at an altitude \(H\), has to drop a food packet for a victim standing on the ground. At what distance from the victim should the food packet be dropped?

    Ans: \(\sqrt{H^{2}+\frac{2u^{2}H}{g}}\).

  5. A projectile is projected from the ground with initial velocity of \(5ms^{-1}\) and at an angle of \(\theta\) with the horizontal, another projectile is fired from another planet with a velocity of \(3ms^{-1}\) at the same angle follows a trajectory which is identical with the trajectory of the projectile fired from the earth. The value of acceleration due to gravitation of the another planet is what? Take: \(\vec{g_{e}}=9.8ms^{-2}\).

    Ans: \(3.5m.s^{-2}\).

  6. An arrow is projected in air. Its time of flight is \(5s\) and range \(200m\). What is the maximum vertical range of the projectile? take: \(\vec{g}=9.8ms^{-2}\).

    Ans: \(31.25m\).

  7. A projectile is projected from the ground with initial velocity \(\vec{u}=u_{o}\hat{i}+v_{o}\hat{j}\), if the acceleration due to gravity \(\vec{g}\) is along vertically downward direction, then find the maximum displacement in x-direction.

    Ans: \(\frac{2u_{o}v_{o}}{g}\).

  8. The velocity of a projectile at the initial point is \(\left(2\hat{i}+3\hat{j}\right)ms^{-1}\). What is its velocity t=at the final point?

    Ans: \(\left(2\hat{i}-3\hat{j}\right)m.s^{-1}\).

  9. An arrow is projected at an angle of \(60^{o}\) above the horizontal with a speed of \(10ms^{-1}\). After some time the direction of velocity makes an angle of \(30^{o}\) above the horizontal. What is the speed of the particle at this instant?

    Ans: \(\frac{10}{\sqrt{3}}m.s^{-1}\).

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