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Does the distance in one second depend on the angle of launch or the speed of the projectile?
Wiki User
∙ 13y ago
The answer is both. Let's say you're shooting a BB gun with an initial velocity of 50 m/s
Say you point it at an angle 36.872° up from horizontal. This gives a 3-4-5 triangle. The initial velocity is directed along the hypotneus of the triangle, and it has components of 40 m/s horizontal, and 30 m/s vertical. If air drag is not taken into account, then after 1 second, the BB would travel 80 meters horizontally. Because gravity is slowing the vertical velocity, the BB will travel less than 30 meters upward in the first second.
Now shoot at a 53.128° angle, and you've flipped the 3-4-5 triangle, so that the initial horizontal velocity is 30 m/s and initial vertical velocity is 40 m/s. Now gravity will be slowing down from 40 m/s initial.
You can use the following equations to solve for velocity and distance due to acceleration:
Vf = V0 + a*t and d = V0*t + (1/2)*a*t2
Vf is final velocity, V0 is initial velocity, a is acceleration [in this case -9.8 m/(sec2) for gravity on Earth], d is distance traveled (vertical distance only), t is time.
For the first angle, @ t = 1 sec, d = (30 m/s)*(1 s) + (1/2)*(-9.8 m/(s2))*(1 s)2 = 25.1 m (vertical), and in 1 sec, it traveled 40 meters horizontal.
For the 2nd angle, vertical distance = 40m - 4.9m = 35.1 m, & 30 m horizontal.
If you went and got a different BB gun with more or less initial velocity, just plug that number in for V0. Of course, real-world evaluations would be different, because air drag is a real factor, and it would act to slow down the horizontal velocity component as well.
Wiki User
∙ 13y ago
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