A projectile is launched at an angle of 45° above the horizontal. What is the ratio of its horizontal range to?
A projectile is launched at an angle of 45° above the
horizontal. What is the ratio of its horizontal range to its
maximum height? How does the answer change if the initial
speed of the projectile is doubled
hello, this is another physics question which it is complicated to me so if there is anyone can help me i will be greatful
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I assume you know these basic equations:
R = v²sin(2θ)/g
h = v²sin²θ /(2g)
So:
R/h = 2sin(2θ)/sin²θ
When θ = 45°,
R/h = 2sin(90°)/sin²(45°)
= 2(1)/(√½)²
= 4
Since the angle is 45: Vyi = Vxi = Vi*sin45= Vi*cos45 where Vyi is the initial vertical component of velocity, Vxi is the initial horizontal component of velocity and Vi is the initial velocity. Use the standard projectile equations for max height and Range
Max height = Vyi²/2g = Vi²*sin²θ/2g
Range = Vi²*sin2θ/g = 2*Vi²*sinθ*cosθ/g
Range/Max Height = 2*Vi²*sinθ*cosθ/g/Vi²*sin²θ/2g = 4cotθ = 4 for θ = 45
You see that Vi drops out of the result which means the ratio is independent of Vi regardless of the launch angle.
If angle is 45* then horizontal dis =2 times vertical.
Independent of intial speed.