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The force due to gravity has no component in horizontal direction. Since gravity is the only force acting on the projectile, this means that the motion in horizontal direction is not accelerated. Therefore, the motion in horizontal direction is an uniform motion. This implies that the component of velocity in x-direction is constant. As such, the position or displacement in x-direction at a given time “t” is :
This equation gives the value of horizontal position or displacement at any given instant.
The displacement of projectile is obtained by vector addition of displacements in x and y direction. The magnitude of displacement of the projectile from the origin at any given instant is :
The angle that displacement vector subtends on x-axis is :
The velocity of projectile is obtained by vector addition of velocities in x and y direction. Since component velocities are mutually perpendicular to each other, we can find magnitude of velocity of the projectile at any given instant, applying Pythagoras theorem :
The angle that the resultant velocity subtends on x-axis is :
Problem : A ball is projected upwards with a velocity of 60 m/s at an angle 60° to the vertical. Find the velocity of the projectile after 1 second.
Solution : In order to find velocity of the projectile, we need to know the velocity in vertical and horizontal direction. Now, initial velocities in the two directions are (Note that the angle of projection is given in relation to vertical direction.):
$$\begin{array}{l}{u}_{x}=u\mathrm{sin}\theta =60\mathrm{sin}60\xb0=60x\frac{\sqrt{3}}{2}=30\sqrt{3}\phantom{\rule{2pt}{0ex}}m/s\\ {u}_{y}=u\mathrm{cos}\theta =60\mathrm{cos}60\xb0=60x\frac{1}{2}=30\phantom{\rule{2pt}{0ex}}m/s\end{array}$$
Now, velocity in horizontal direction is constant as there is no component of acceleration in this direction. Hence, velocity after "1" second is :
$$\begin{array}{l}{v}_{x}={u}_{x}=30\sqrt{3}\phantom{\rule{2pt}{0ex}}m/s\end{array}$$
On the other hand, the velocity in vertical direction is obtained, using equation of motion as :
$$\begin{array}{l}{v}_{y}={u}_{y}-gt\\ \Rightarrow {v}_{y}=30-10x1\\ \Rightarrow {v}_{y}=20\phantom{\rule{2pt}{0ex}}m/s\end{array}$$
The resultant velocity, v, is given by :
$$\begin{array}{l}v=\sqrt{({{v}_{x}}^{2}+{{v}_{y}}^{2})}\\ \Rightarrow v=\sqrt{\{{\left(30\sqrt{3}\right)}^{2}+{\left(20\right)}^{2}\}}=\sqrt{(900x3+400)}=55.68\phantom{\rule{2pt}{0ex}}m/s\end{array}$$
Equation of projectile path is a relationship between “x” and “y”. The x and y – coordinates are given by equations,
$$\begin{array}{l}y={u}_{y}t-\frac{1}{2}g{t}^{2}\\ x={u}_{x}t\end{array}$$
Eliminating “t” from two equations, we have :
For a given initial velocity and angle of projection, the equation reduces to the form of $y=Ax+B{x}^{2}$ , where A and B are constants. The equation of “y” in “x” is the equation of parabola. Hence, path of the projectile motion is a parabola. Also, putting expressions for initial velocity components ${u}_{x}=u\mathrm{cos}\theta \mathrm{and}{u}_{y}=u\mathrm{sin}\theta $ , we have :
Some other forms of the equation of projectile are :
A projectile with initial velocity 2 i + j is thrown in air (neglect air resistance). The velocity of the projectile before striking the ground is (consider g = 10 $\phantom{\rule{2pt}{0ex}}m/{s}^{2}$ ) :
(a) i + 2 j (b) 2 i – j (c) i – 2 j (d) 2 i – 2 j
The vertical component of velocity of the projectile on return to the ground is equal in magnitude to the vertical component of velocity of projection, but opposite in direction. On the other hand, horizontal component of velocity remains unaltered. Hence, we can obtain velocity on the return to the ground by simply changing the sign of vertical component in the component expression of velocity of projection.
v = 2 i - j
Hence, option (b) is correct.
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