**Projectile Motion Problems:**

The problems based on the projectile motion can be of following types:

1. The initial height of the projectile is lower than the final height of the projectile. In the given figure, the initial height of projectile is A and final height is B, which is a higher point relative to A.

2. The initial height of the projectile is greater than the final height of the projectile. The initial height of projectile i.e. A is higher with respect to the final height of projectile i.e. B.

3. The initial and final height of the projectile is same. The initial heightÂ A of the projectile is same as the final height B of the projectile as shown in the figure.

Let us now solve a few projectile motion problems based on these cases.

Problem 1: An angry bird is launched with an initial velocity of \(10âˆš2 m/s\)

Solution: From the data given in the problem,

Component of velocity in x direction,

\(v_{ox}~=~v_o~cos~Î¸~=~10âˆš2~cos~45^{\circ}~=~10 ~m/s\)

Component of velocity in y-direction,

\(v_{oy}~=~v_o~sin~Î¸~=~10âˆš2~sin~45^{\circ}~=~10~ m/s\)

As there is no acceleration in the horizontal direction and the velocity in the horizontal direction is constant. Also, the distance traveled in horizontal direction or range is given as \(10 m\)

Since, \(Range~=~velocity~Ã—~time\)

\(â‡’~10~=~10t\)

\(â‡’~t~=~1~ sec\)

This is the time required for the bird to reach the wooden platform, which is at a height of \(15 ~m\)

Now, will the bird even hit the pig? Let us try to figure that out.

The distance traveled by the projectile in the y-direction is at the end of 1 sec will be:

\(s_y~=~ v_{oy}~ t~ +~\frac{1}{2}~ a_y t^2\)

\(â‡’~s_y~=~10(1)~-~\frac{1}{2}~ 10(1)^2\)

\(â‡’~s_y~=~5 m\)

So, the bird is at a height of 5m at the end of 1sec and the pig is sitting at a height of 15 m. Therefore, it will miss the pig and he is going to leave. The motion of the bird will be as shown in figure.

Problem 2: A missile is being launched from a cliff of a given height as shown in the figure given below. Calculate the range of the projectile. Assume the acceleration due to gravity is \(10 ~m/s^2\)

Solution: Such projectile motion problems can be solved easily by resolving the initial velocity into vertical and horizontal components. This can be considered as the sum of two one-dimensional motions.

Resolving the initial velocity, we have

Since the time of flight depends upon the component of velocity in the vertical direction. Hence, using the equations of motion in y-direction and considering upward displacement as positive in y-direction and forward displacement positive in x-direction, we have

\(s_y~=~ v_{oy}~ t~ +~\frac{1}{2}~ a_y t^2\)

\(â‡’-45~=~40t~+~\frac{1}{2}~ (10) ~(t)^2\)

\(â‡’~5t^2~-~40t~-~45~=~0\)

\(â‡’~t^2~-~8t~-~9~=~0\)

\(â‡’~t~=~9~ sec\)

\(t~=~-1\)

The time,Â taken by the projectile to reach back to the same height as point A is \(9 ~sec\)

Now, \(Range~=~Velcoity~Ã—~time\)

\(â‡’~R~=~(40~Ã—~9)~m\)

\(â‡’~R~=~360 ~m\)

Thus, the range of projectile is \(360~ m\)

Problem 3: In a cricket match, batsman hits the ball at a velocity of \(10âˆš2 ~ m/s\)

Solution: The initial velocity can be resolved as

According to the given problem:

\(v_{oy}~=~10\)

\(a~=~-10 ~m/s\)

\(s_y~=~3 m\)

According to equations of motion in a plane:

\(s_y~= ~v_{oy}~ t~ +~\frac{1}{2}~ a_y ~t^2\)

Substituting the values we have;

\(3~=~10(t)~-~\frac{1}{2} ~(10) ~t^2\)

\(5t^2~-~10t~+~3~=~0\)

Solving for \(t\)

\(t~=~1~Â±~\frac{\sqrt{40}}{10} ~sec\)

Thus, the projectile motion problems may seem to appear as different but they can be easily solved using the correct approach.’