Three factors which determine the horizontal distance traveled: how does the velocity changes in each direction? An object is projected horizontally. what are three factors which determine the...
Three factors which determine the horizontal distance traveled: how does the velocity changes in each direction?
An object is projected horizontally. what are three factors which determine the horizontal distance traveled, and describe how each factor affects this distance, how does the velocity changes in each direction?
In typical simplified projectile motion, there are two axes of movement: vertical (y-axis) and horizontal (x-axis). The object flying through the air is in free-fall, meaning the net force acting on the ball is due to only the interaction of the Earth and object through gravity (i.e. a = 9.8 m/s^2 vertically downward).
Since there is no net force horizontally, the horizontal velocity does not change.
So the range of a projectile is derived from the equation for objects moving at constant velocity:
`\Deltax = v_x*\Deltat`
∆t, however, has some limiting factors: look to the equation for accelerated motion:
`y_(f) = 1/2a*\Deltat^2+v_(i,y)*\Deltat+y_i`
y_f is the height where the ball strikes the ground, and y_i is the initial height. Combining these two terms into one is ∆y, the change in height. If ∆y=0, then the ball lands at the same height as it was launched. The amount of time to land can be varied, and thus the range varied, if the change in height is differed.
since `a = (\Deltav)/(\Deltat) and v_f^2 = v_i^2+2a\Deltay`
`\Deltat = (+-sqrt(v_(i,y)^2+2a\Deltay)-v_(y,i))/a_y`
Putting it together we get
`v_x and v_y,i` are components of the initial velocity vector `vec(v)` .
Which has a magnitude `v` and a direction `\theta`
Thus: three factors that the range is based on:
The initial velocity, angle of launch, and the change in height. Of course, the fourth factor (that doesn't change for most problems) is the acceleration due to gravity.
An example: ∆y = -10 m, a = -9.8 m/s, v_i = 20 m/s at 30°
`v_(x) = v_i*cos(30) = 17.3 m/s`
`v_(i,y)=v_i*sin(30) = 10 m/s`
`\Deltax =17.3 m/s*(+-sqrt((10 m/s)^2+2(-9.8 m/s^2)(-10m))-10m/s)/(-9.8m/s)`
Since the projectile will be heading downward at the end of the trip, we choose the negative result of the squareroot...
∆x = 48.0 m