what minimum initial velocity must aprojectile have to reach atarget at a distance of 25m away? (g: 10m/s)
do you know max range is 45 degrees above horizontal ? If so
speed S
u = horizontal speed = s cos 45
Vi = initial vertical speed = S sin 45
time in air = 25/u
half time in air = time rising = 12.5/u = t
v = Vi - 10 t
v = 0 at top of rise phase
so
Vi = 10 t
Vi = 10 (12.5/u)
S sin 45 = 125 /S cos 45
S^2 sin 45 cos 45 = 125
but sin 45 = cos 45 = sqrt 2/2
so
S^2 (1/2)= 125
S = sqrt 250 = 15.8 m/s
http://www.wired.com/2010/09/maximum-range-in-projectile-motion/
To find the minimum initial velocity of a projectile required to reach a target at a certain distance, we can use the kinematic equations of motion. In this case, the relevant equation is the horizontal distance formula:
d = v0 * t
Where:
- d is the horizontal distance (25m)
- v0 is the initial velocity we want to find
- t is the time it takes for the projectile to reach the target
Since the motion is in the horizontal direction, the acceleration (g) due to gravity does not affect the horizontal velocity. Therefore, we can consider the motion as uniform (constant velocity).
In this case, the projectile will be in the air for the same amount of time it takes to hit the target. Using the equation for time:
t = 2 * (d / v0)
Substituting the value of d (25m) into the equation, we can rewrite it as:
t = 2.5 / v0
Since we want to find the minimum initial velocity, we need to consider the worst-case scenario, where the projectile is launched at the minimum possible angle (i.e., 45 degrees). At this angle, the vertical and horizontal components of velocity are equal, so the initial velocity can be calculated using:
v0 = v * sqrt(2)
Where v is the velocity component in one direction (either vertical or horizontal).
In this case, we need to find v0. Considering the horizontal component of the initial velocity:
v0_horizontal = v_horizontal * sqrt(2)
Since the motion is in a horizontal direction, there is no vertical component of velocity initially (v_vertical = 0). Thus, v_horizontal = v0_horizontal.
Now, substituting v0_horizontal into the time equation:
t = 2.5 / (v0_horizontal * sqrt(2))
To find the minimum initial velocity, we need to find the smallest possible value of v0_horizontal that satisfies the equation.
By rearranging the equation, we get:
v0_horizontal * sqrt(2) = 2.5 / t
v0_horizontal = (2.5 / t) / sqrt(2)
Finally, substituting t = 2.5 / v0 into the equation:
v0_horizontal = (2.5 / (2.5 / v0)) / sqrt(2)
Simplifying,
v0_horizontal = v0 / sqrt(2)
Therefore, the minimum initial velocity of the projectile required to reach a target at a distance of 25m away is equal to 25 / sqrt(2) m/s, or approximately 17.68 m/s.