At what angle should a body be projected with a velocity 24 ms^{-1} just to pass over the obstacle 16 m high at a horizontal distance of 32 m . Assume g = 10 ms^{-1}?

**Solution:**

## what angle should

If point of projection is taken as the origin of the coordinate system, the projected body must pass through a point having coordinates (32 m, 16 m) as the following figure.

Suppose u be the initial velocity of the projectile. If θ be the angle of projection, then

Horizontal component of initial velocity u_{x }= u cos θ.

And vertical component of initial velocity u_{y} = u sin θ.

Suppose the body passes through point P at time t. then horizontal distance covered is x = 32 m i.e., 32 = x = (u cos θ ) t or 32 = (24 cos θ) t …….(1)

*Similarly, vertical distance covered is y = 16 m i.e.,*

To better understanding learn the equation of motion.

16 = y = (u sin θ)t-1/2gt^{2}

Or 16 = (24 sin θ)t – 1/2x 10x t^{2} …..(2)

From equation (1) we get : t = 32/(24 cos θ)

*Putting this value of t in equation (2), we have:*

16 = (24 sin θ) 32/(24 cos θ) – ½ x 10x (32/24cos θ)^{2}

Or 16 = 32 tan θ – 5 x 16/9cos^{2} θ

Or 1 = 2 tan θ – 5/9 sec^{2} θ or 9 = 18 tan θ – 5(1+tan^{2} θ)

Or 5 tan^{2} θ – 18 tan θ + 14=0

Solving above equation: tan θ = [18 ±√(18)^{2} -4 x 5 x 14]/10= 2.462 or 1.137

So, Θ = 67⁰54’ or 48⁰40’