For a given initial projectile speed Vo, calculate what launch angle A gives the longest range R. Show your work, don't just quo

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makkiz

1 month ago

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For a given initial projectile speed Vo, calculate what launch angle A gives the longest range R. Show your work, don't just quo

te a number.

Physics

1 answer:

ValentinkaMS [3.4K] · Answer 1 · 2026-04-10T12:28:48+03:00

The ideal launch angle of 45° for achieving the greatest horizontal distance is only applicable when the starting height matches the final height.

In this scenario, you can demonstrate it as follows: 

the initial velocity is Vo 
the launch angle is α 

the initial vertical velocity is 
Vv = Vo×sin(α) 

horizontal velocity becomes 
Vh = Vo×cos(α) 

the total flight duration is the period required to return to a height of 0 m, thus 
d = v×t + a×t²/2 
where 
d = distance = 0 m 
v = initial vertical velocity = Vv = Vo×sin(α) 
t = time =? 
a = gravitational acceleration = g (= -9.8 m/s²) 
therefore 
0 = Vo×sin(α)×t + g×t²/2 
0 = (Vo×sin(α) + g×t/2)×t 
t = 0 (obviously, the projectile is at height 0 m at time = 0s) 
or 
Vo×sin(α) + g×t/2 = 0 
t = -2×Vo×sin(α)/g 

Now let's examine the horizontal distance. 
r = v × t 
where 
r = horizontal range =? 
v = horizontal velocity = Vh = Vo×cos(α) 
t = time = -2×Vo×sin(α)/g 
therefore 
r = (Vo×cos(α)) × (-2×Vo×sin(α)/g) 
r = -(Vo)²×sin(2α)/g 

To find the extreme points of r (max or min) with respect to α, the first derivative of r with regards to α must be determined and set to 0. 

dr/dα = d[-(Vo)²×sin(2α)/g] / dα 
dr/dα = -(Vo)²/g × d[sin(2α)] / dα 
dr/dα = -(Vo)²/g × cos(2α) × d(2α) / dα 
dr/dα = -2 × (Vo)² × cos(2α) / g 

As Vo and g are constants that are not equal to 0, the only solution for dr/dα to equal 0 is when 
cos(2α) = 0 
2α = 90° 
α = 45°