A cricket ball is thrown at a speed of \(28 ms^{-1}\) in a direction \(30^{o}\) above the horizontal direction. Calculate: (a) Vertical Range. (ii) Time of Flight. (iii) Horizontal Range.
Ans:(a) 10m, (b) 2.9s and (c) 69.3m.
A stone is thrown at an angle of \(30^{o}\) with the vertical. If, the horizontal component of its velocity is \(19.6 ms^{-1}\). Find the maximum height and hoizontal range attained by the stone?
Ans: \(H=58.8m\) and \(R=135.8m\).
Show that: the projection angle \(\theta\) for a projectile launched from origin is given by \(\theta=tan^{-1}\left(\frac{4H}{R}\right)\). Where; R=Horizontal Range, and H= Vertical Range.
A projectile has a range of \(60m\) and reaches maximum height of \(12m\). Calculate: the angle at which projectile is fired and initial velocity of the projection?
Ans:(i) \(38^{o}\) and (ii) \(24.8m.s^{-1}\).
A person stands at \(39.2m\) far from a house and throw a stone stone which just passes through a window \(19.6m\) above the gound. Calculate: the velocity of projection of the stone?
Ans: \(u=27.72m.s^{-1}\).
For, angles of projection which exceed or fall-short of \(45^{o}\) by an equal amount, the ranges are equal. Prove the statement.
Show that: a given gun will shot 3 times as high when elevated at an angle \(60^{o}\) as when fired at an angle of \(30^{o}\) but will carry the same distance on a horizontal plane.
If, \(R\) be the horizontal range for \(\theta\) inclination and \(H\) be the maximum height reached by the projectile. Show that: the maximum range is given by, \(\frac{R^{2}}{8H}+2H\).
Two projectile of same mass having their maximum kinetic energy in the ratio \(4:1\) and ratio of their maximum height is also \(4:1\). Then, what is the ratio of their range?
Ans: \(R_{1}:R_{2}=4:1\).
The expression of the trajectory of a particle is given as, \(y=\left(px-qx^{2}\right)\), Where; y and x are respectively are the vertical and horizontal displacements and p and q are constant. What is the expression of time of the flight of the projectile?
Ans: \(T_{flight}=p\sqrt{\frac{2}{qg}}\).
A projectile is given an initial velocity of \(\vec{u}=\left(\hat{i}+2\hat{j}\right) ms^{-1}\). Where; \(\hat{i}\) along the ground and \(\hat{j}\) is along the vertical. If, g is taken as \(10 ms^{-20}\), then what is the equation of trajectory?
Ans: \(y=\left(2x-5x^{2}\right)\).
A projectile is thrown with initial velocity of, \(\vec{u}=\left(p\hat{i}+q\hat{j}\right) ms^{-1}\). If, the range of the projectile is doubled, the maximum height attained by the projectile?, then what is the relation between p and q?
