Projectile motion  (Page 4/6)

The time to reach maximum height in this case is :

$$\begin{array}{l}\Rightarrow t=\frac{u_y}{g}=\frac{30}{10}=3s\end{array}$$

The data in the table confirms this. Further, we know that vertical motion is independent of horizontal motion and time of flight for vertical motion is equal for upward and downward journey. This means that total time of flight is 2t i.e. 2x3 = 6 seconds. We must, however, be careful to emphasize that this result holds if the point of projection and point of return to the surface are on same horizontal level.

There is yet another interesting feature that can be drawn from the data set. The magnitude of vertical velocity of the projectile (30 m/s) at the time it hits the surface on return is equal to that at the time of the start of the motion. In terms of velocity, the final vertical velocity at the time of return is inverted initial velocity.

Displacement

The displacement in vertical direction is given by :

This equation gives vertical position or displacement at a given time. It is important to realize that we have simplified the equation $\Delta y=y_2-y_1=u_{y}t-\frac{1}{2}g{t}^2$ by selecting origin to coincide by the point of projection so that,

$$\begin{array}{l}\Delta y=y_2-y_1=y\left(\mathrm{say}\right)\end{array}$$

Thus, “y” with this simplification represents position or displacement.

The equation for position or displacement is a quadratic equation in time “t”. It means that solution of this equation yields two values of time for every value of vertical position or displacement. This interpretation is in fine agreement with the motion as projectile retraces all vertical displacement as shown in the figure.

Time of flight

The time taken to complete the journey from the point of projection to the point of return is the time of the flight for the projectile. In case initial and final points of the journey are on the same horizontal level, then the net displacement in vertical direction is zero i.e. y = 0. This condition allows us to determine the total time of flight “T” as :

$$\begin{array}{l}y=u_{y}T-\frac{1}{2}g{T}^2\\ \Rightarrow 0=u_{y}T-\frac{1}{2}g{T}^2\\ \Rightarrow T(u_y-\frac{1}{2}gT)=0\\ \Rightarrow T=0\mathrm{or}T=\frac{2u_y}{g}\end{array}$$

T = 0 corresponds to the time of projection. Hence neglecting the first value, the time of flight is :

We see that total time of flight is twice the time projectile takes to reach the maximum height. It means that projectile takes equal times in "up" and "down" motion. In other words, time of ascent equals time of descent.