Question

A projectile is shot from the ground at an angle of `pi/12` and a speed of `2/5 m/s`. When the projectile is at its maximum height, what will its distance from the starting point be?

Answer 1

Please see the explanation

Explanation:

Assume the projectile is launched with a speed `v_0` at an angle `\theta_0`.

Time of Flight: `T= (2v_0\sin\theta_0)/g`

Range of the projectile: `R = (2v_0^2\sin\theta_0\cos\theta_0)/g = (v_0^2\sin2\theta_0)/g`

Height of the projectile: `H = (v_0^2\sin^2\theta_0)/(2g)`

Given that : `v_0 = 2/5 ms^{-1}; \qquad \theta_0 = \pi/12`
Now substitute the numerical values of `v_0`, `\theta_0` and `g`, calculate the numerical values of `R` and `H`.

When the projectile is at maximum height, its coordinates are `(R/2, H)`

So its distance from the origin is ; `r = \sqrt{(R/2)^2 + H^2} = (\sqrt{R^2+4H^2})/2`

From the numerical values of `R` and `H`, calculated earlier, evaluate `r`.