# Projectile Motion- Jumping off Cliffs

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 $10âˆš2 m/s$ at an angle of $45^{\circ}$ as shown. Determine the time in which it will hit the pig on the wooden platform. Assume the acceleration due to gravity is $10 m/s^2$.

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 $15 ~m$ from the ground as given.

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

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

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

$â‡’~5t^2~-~40t~-~45~=~0$

$â‡’~t^2~-~8t~-~9~=~0$

$â‡’~t~=~9~ sec$ and $t~=~-1~ sec$

$t~=~-1$ implies that the time taken for a ball launched at position A to reach at position B with the velocity of $40~ m/s$ is $1 ~sec$.

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 $10âˆš2 ~ m/s$ at an angle of $45^{\circ}$ in order to score a six. A bowler can catch a ball, which is at a height of 3 m with respect to the ground. Determine the time in, which bowler will catch the ball.

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 $t$ we get,

$t~=~1~Â±~\frac{\sqrt{40}}{10} ~sec$

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

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