<< Chapter < Page | Chapter >> Page > |
The vertical component velocity of the projectile reduces to zero as motion decelerates while going up against the force due to gravity. The point corresponding to this situation, when vertical component of velocity is zero, ${v}_{y}=0$ , is the maximum height that a projectile can reach. The projectile is accelerated downward under gravity immediately thereafter. Now, considering upward motion in y-direction,
$$\begin{array}{l}{v}_{y}={u}_{y}-gt\\ \Rightarrow 0={u}_{y}-gt\\ \Rightarrow \mathrm{t}=\frac{{u}_{y}}{g}\end{array}$$
Now, using equation for displacement in the vertical direction, we can find out the vertical displacement i.e. the height as :
$$\begin{array}{l}y={u}_{y}t-\frac{1}{2}g{t}^{2}\\ \Rightarrow H={u}_{y}\frac{{u}_{y}}{g}-\frac{1}{2}{\frac{{u}_{y}}{g}}^{2}\\ \Rightarrow H=\frac{{{u}_{y}}^{2}}{g}-\frac{{{u}_{y}}^{2}}{2g}\\ \Rightarrow H=\frac{{{u}_{y}}^{2}}{2g}\end{array}$$
Putting expression of component velocity ( ${u}_{y}=u\mathrm{sin}\theta $ ), we have :
We can also obtain this expression, using relation $\begin{array}{l}{{v}_{y}}^{2}={{u}_{y}}^{2}+2gy\end{array}$ . Just like the case of the time of flight, we see that maximum height reached by the projectile depends on both initial speed and the angle of projection (θ). Greater the initial velocity and greater the angle of projection from horizontal direction, greater is the height attained by the projectile.
It is important to realize here that there is no role of the horizontal component of initial velocity as far as maximum height is concerned. It is logical also. The height attained by the projectile is purely a vertical displacement; and as motions in the two mutually perpendicular directions are independent of each other, it follows that the maximum height attained by the projectile is completely determined by the vertical component of the projection velocity.
The maximum height that a projectile, thrown with an initial speed " ${v}_{0}$ ", can reach :
$$\left(a\right)\frac{{v}_{0}}{2g}\phantom{\rule{1em}{0ex}}\left(b\right)\frac{{v}_{0}^{2}}{2g}\phantom{\rule{1em}{0ex}}\left(c\right)\frac{{v}_{0}^{2}}{g}\phantom{\rule{1em}{0ex}}\left(d\right)\frac{{v}_{0}}{g}$$
The question only specifies speed of projection - not the angle of projection. Now, projectile rises to greatest maximum height for a given speed, when it is thrown vertically. In this case, vertical component of velocity is equal to the speed of projection itself. Further, the speed of the projectile is zero at the maximum height. Using equation of motion, we have :
$$0={v}_{0}^{2}-2gH$$ $$\Rightarrow H=\frac{{v}_{0}^{2}}{2g}$$
The assumption for the maximum height as outlined above can also be verified from the general formula of maximum height of projectile as given here :
$$H=\frac{{u}_{y}^{2}}{2g}=\frac{{u}^{2}{\mathrm{sin}}^{2}\theta}{2g}=\frac{{v}_{0}^{2}{\mathrm{sin}}^{2}\theta}{2g}$$
The numerator of the above is maximum when the angle is 90°.
$$\Rightarrow H=\frac{{v}_{0}^{2}}{2g}$$
We should note that the formula of maximum height for a projectile projected at certain angle represents the maximum height of the projectile for the given angle of projection and spped. The question here, however, refers to maximum height for any angle of projection at given speed of projection. As such, we should consider an angle of projection for which projectile reaches the greatest height. This point should be kept in mind.
Hence, option (b) is correct.
The horizontal range is the displacement in horizontal direction. There is no acceleration involved in this direction. Motion is an uniform motion. It follows that horizontal range is transversed with the horizontal component of the projection velocity for the time of flight (T). Now,
Notification Switch
Would you like to follow the 'Physics for k-12' conversation and receive update notifications?