A projectile on level ground is launched with a certain speed vi and has a downrange distance of vx what is its new downrange distance if it is launched with a speed of 4vi instead?
To find the new downrange distance when the projectile is launched with a speed of 4vi instead of vi, we need to consider the relationship between the initial velocity and the downrange distance.
The downrange distance, vx, is directly related to the initial velocity, vi, of the projectile. This relationship can be described by the equation:
vx = vi * t,
where vx is the downrange distance, vi is the initial velocity, and t is the time taken by the projectile to travel the downrange distance.
Since the downrange distance remains the same, vx, and only the initial velocity changes to 4vi, we can rewrite the equation as:
vx_new = (4vi) * t_new.
Next, we need to determine the relationship between the time of flight for the projectile when launched with vi and when launched with 4vi.
The time of flight (also known as the total time) for a projectile can be determined using the equation:
t = (2 * vi * sinθ) / g,
where θ is the angle of projection and g is the acceleration due to gravity.
Assuming the angle of projection remains the same for both cases (vi and 4vi), we can use this equation to find t_new:
t_new = (2 * (4vi) * sinθ) / g
= (8vi * sinθ) / g.
Now, substituting this value of t_new into the equation for vx_new, we get:
vx_new = (4vi) * t_new
= (4vi) * ((8vi * sinθ) / g)
= 32vi² * (sinθ / g).
Therefore, the new downrange distance, vx_new, when the projectile is launched with a speed of 4vi is given by 32vi² * (sinθ / g).