💯An Ingelious Way to Understand Kinematics Projectile Motion Questions

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Kinematics Projectile Motion

Here’s an “ingelious” way to understand JC A Level H2 Physics projectile motion related to kinematics.
Let’s try to derive the optimum angle to throw a javelin for maximum range, assuming no air resistance:

The distance travelled by a projectile is always determined by its time of flight t .
Its time of flight t is in turn determined by its initial vertical velocity v_y  .
Note t is the same for both horizontal and vertical motion.
To find its time of flight, let the vertical displacement s_y be 0 (when the javelin returns back to ground)
Set upwards as positive,
s_y = u_yt + \frac{1}{2} a_yt^2

It is good practice not to change the sign of the formula, but use (-g)
0 = usin \theta t + \frac{1}{2} (-g)t^2
0 = usin \theta t - \frac{1}{2} gt^2
0 = (usin \theta - \frac{1}{2} gt)(t)
t = 0 \, (reject) \, or \, t = \frac{2usin\theta}{g}

Since range s_x = u_x t
s_x = ucos\theta t
s_x = \frac{2u^2sin\theta cos\theta}{g}
s_x = \frac{u^2sin2\theta}{g}

Since s_x, u, g are constants,
To maximise s_x, sin2\theta = 1  ,
\therefore \theta = 45^{\circ} ,

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