Question

Derive the expression for the time of flight and maximum height of projectile​

Answer 1

$\huge\underline\red{Answer}$

Let a projectile move with a velocity u which is inclined with the horizontal at angle of θ. The velocity after time 't' will be

${v = v_xî + v_yj}$

${v=ucosθî+(usinθ−gt)j}$

${(Since:v_y = usinθ−gt)}$

At maximum height the projectile will only have horizontal component that is

${v_x = ucosθ}$

${v_y2 −u _y2 = 2ay}$

${v_y =0 (at\:max\:height\:H)}$

${u_y = usinθ}$

${ay= −gH}$

Putting these values,

$0 = (usinθ)^2 −2gH$

$\huge{H = \frac{u^2sin^2θ} {2g}}$