Question

Which combination of initial horizontal velocity, (vh) and initial vertical velocity, (vv) results in the greatest horizontal range for a projectile over level ground? [neglect friction.] 1. 2. 3. 4?

Answer 1

If an object is projected with vertical speed given as

$v_v$

now the time of flight of the object that time in which it comes back on ground

$\Delta y = v_v t + \frac{1}{2}(-g)t^2$

now here we will have

$0 = v_v t - \frac{1}{2}gt^2$

$t = \frac{2v_v}{g}$

now the range of projectile is given as

$R = horizontal\: speed \times time$

$R = v_h(\frac{2v_v}{g}$

now here we know that

$v_v = v_0 sin\theta$

$v_h = v_0 cos\theta$

now the range is given as

$R = \frac{2(vsin\theta)(vcos\theta)}{g}$

$R = \frac{v^2sin2\theta}{g}$

now in order to have maximum range we can say

$sin2\theta = 1$

$2\theta = 90^0$

so we will have

$\theta = 45 ^0$

so now we can say

$v_v = v_h = \frac{v_0}{\sqrt2}$

so both speed must be same to have maximum horizontal range