Solving Projectile Motion Problems

There are four types of projectile motion as we learned in the physics class. They are, for example, the projectile motion one, object falls from top to the horizontal. In this diagram, the object is moving horizontally with a down force that pulls the object down to the ground. This force is know as gravity, which is 9.8m/s square.

Projectile Motion 1

Solving this type of question, usually the x value (distance travelled) is given, as well as the vx value (horizontal velocity). The common questions are: asking for time travelling the x distance; knowing the time, asking for height (dy).




Example 1:  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.



This is a typical question for projectile motion 1. The commonly used equations are at = v2 - v1, and d = v1t + 1/2gt square.

i) for finding the time, we substitute the values given and used equation
d = v1t + 1/2gt square.
Since d = 2000, g (gravity, constant value one Earth) = 9.81, and v1 (start with 0 metre per second) = 0m/s

The new equation will be: 2000 = 0(t) + 1/2(9.81)(t) square; following the equation solving process, we will get a final answer of 20.20 second.
ii)We can find the velocity at the time of strike with ground by calculating component velocities at that instant in the  perpendicular directions and finding the resultant (composite) velocity as : vf = vy square + vx square. The answer will be 223.60 m/s
iii)Horizontal distance traveled by the object: time x horizontal velocity
20.20 x 100 = 2000 m

Projectile Motion 2

All questions about projectile motion 2 are given either the initial velocity and angle or vertical velocity and horizontal velocity. If initial velocity and angle are given, using cos θ multiplying initial velocity to find horizontal velocity, and using sin θ multiplying initial velocity to find vertical velocity.

Three common questions for this type of projectile motion are: asking for range, asking for height, and asking for travelling time.

Example 2: A baseball player leads off the game and hits a long home run. The ball leaves the bat at an angle of 30.0 degrees from the horizontal with a velocity of 40.0 m/s. How far will it travel in the air?

Using the values given, we know the initial velocity is 40.0 m/s and the angle of the motion is 30.0 degrees.  We then use cosθ to get horizontal velocity (vx), which is 36.64 m/s, and the vertical velocity (vy1), which is 20 m/s.

Since we all know distance equals time multiplying speed, and horizontal velocity is a constant value. We need to calculate the time first in order to get the distance travelled. There is a simply equation used to find time. t = 2sinθ/g
or following a more complicated method which is to calculate the time used to reach the vertical velocity from 0 m/s. In this case, I chose to use the second method. The vertical velocity is 20 m/s, and we divide it by 9.81, which is the gravity, we get 2.04 second for the object to reach the highest point and the vertical velocity. Use the time we got multiplying by 2, we will get the total time for travelling the entire distance, which is 4.08 seconds in this case. 4.08 times 34.64 (horizontal distance), we get the range, which is 141 m.