5.4 Projectile motion types  (Page 3/5)

Problem : A plane flying at the speed of 100 m/s parallel to the ground drops an object from a height of 2 km. Find (i) the time of flight (ii) velocity of the object at the time it strikes the ground and (iii) the horizontal distance traveled by the object.

Solution : The basic approach to solve the problem involves consideration of motion in two mutually perpendicular direction. Here, we consider a coordinate system with the point of release as the origin and down ward direction as the positive y-direction.

(i) Time of flight, T

In vertical direction :

$$\begin{array}{l}{u}_y=0,a=10m/s^2,y=2000m,T=?\end{array}$$

Using equation, $y=u_{y}T+\frac{1}{2}g{T}^2$ , we have :

$$\begin{array}{l}\Rightarrow y=\frac{1}{2}g{t}^2\\ \Rightarrow T=\sqrt{\left(\frac{2y}{g}\right)}\\ \Rightarrow T=\sqrt{\left(\frac{2x2000}{10}\right)}=20s\end{array}$$

(ii) Velocity at the ground

We can find the velocity at the time of strike with ground by calculating component velocities at that instant in the two mutually perpendicular directions and finding the resultant (composite) velocity as :

$$\begin{array}{l}v=\sqrt{({v_x}^2+{v_y}^2)}\end{array}$$

Initial vertical component of initial velocity (uy) is zero and the object is accelerated down with the acceleration due to gravity. Hence,

$$\begin{array}{l}{v}_y=u_y+gt=0+gt=gt\end{array}$$

The component of velocity in the horizontal direction remains unchanged as there is no acceleration in this direction.

$$\begin{array}{l}{v}_x=u_x=u\end{array}$$

$$\begin{array}{l}\Rightarrow v=\sqrt{({v_x}^2+{v_y}^2)}=\sqrt{(u^2+g^2{t}^2)}\end{array}$$

Putting values,

$$\begin{array}{l}\Rightarrow v=\sqrt{({100}^2+{10}^{2}x{20}^2)}=\sqrt{50000}=100\sqrt{5}m/s\end{array}$$

(iii) Horizontal distance traveled

From consideration of uniform motion in horizontal direction, we have :

$$\begin{array}{l}x=u_{x}t=ut\end{array}$$

Putting values,

$$\begin{array}{l}\Rightarrow x=R=100x20=2000m\end{array}$$