Exam Tips: Solving projectile motion

Projectile motion question is always one of the hard question to score. Reason is mainly due to weak foundation in kinematics. There are only two different cases that question can base on., i.e. case with constant velocity and case with constant acceleration.

Projectile motion involves the two cases in one single question, whereby vertical component of the projectile is a case of constant acceleration, and horizontal component is a case of constant velocity.
Reason being, there is only one force acting, i.e. the gravity in the vertical direction.

Let's look at one example question and the suggested steps of solution:

Example question: 9702/21/M/J/2016 Question 2(b)

A ball is thrown from a point P with an initial velocity u of 12 m s^-1 at 50° to the horizontal, as illustrated in Fig. 2.1.

The ball reaches maximum height at Q.
Air resistance is negligible.
Show that the maximum height reached by the ball is 4.3 m.

Steps of solution:
1. Decide component to be used: vertical components --> constant acceleration
2. Assumption of direction         : upward is positive
3. Constant acceleration, so, SUVAT method:
         
                              S = to prove ,
                              U = 9.2 ,
                              V = 0.0 ,
                              A = g = - 9.81,
                                t =?

4. Equation to use:     V² = U² + 2gS, because t is not given.
    Substitute         :   0.0² = 9.2² + 2(-9.81) S
                                      S = 4.31