Your projectile launching system is partially jammed. It can only launch objects with an initial vertical velocity of 42.0 m/s, though the horizontal component of the velocity can vary. You need your projectile to land 214 m from its launch point. What horizontal velocity do you need to program into the system?
the height is
h(t) = 42t - 4.9t^2 = t(42-4.9)
h=0 when t=8.57
so, after 8.57 seconds, the horizontal distance traveled is 214m. Since the horizontal component dies not change (much)
Vx = 214m/8.57s = 25 m/s
How did you come up with 4.9t^2?
To find the required horizontal velocity for the projectile launcher, we can use the following equation:
Range = horizontal velocity * time
In this case, the range is given as 214 m.
The time of flight for a projectile can be determined using the vertical motion equation:
Vertical Displacement = Initial Vertical Velocity * Time - (1/2) * Acceleration Due to Gravity * Time^2
Since the vertical displacement is zero at the end of the motion, we can solve this equation for time:
0 = (42.0 m/s) * Time - (1/2) * (9.8 m/s^2) * Time^2
Simplifying the equation gives us:
(1/2) * (9.8 m/s^2) * Time^2 = (42.0 m/s) * Time
Simplifying further:
4.9 m/s^2 * Time^2 - 42.0 m/s * Time = 0
Solving this quadratic equation, we get two possible values for time: t1 = 0 s and t2 = 8.622 s.
Since the projectile will be in the air for a positive time, t1 = 0 s is not valid. Therefore, the time of flight is 8.622 s.
Now we can substitute the time of flight into the range equation:
214 m = horizontal velocity * 8.622 s
Solving for the required horizontal velocity yields:
horizontal velocity = 214 m / 8.622 s ≈ 24.81 m/s
Therefore, you need to program a horizontal velocity of approximately 24.81 m/s into the system to achieve a range of 214 m.