Final answer:
The calculated horizontal distance of the projectile, which should be 55 meters, based on the projectile motion formulas using an initial velocity of 25 m/s at a 30-degree launch angle. So, none of the given options are correct.
Step-by-step explanation:
To calculate the horizontal distance traveled by a projectile launched at an initial velocity of 25 m/s at an angle of 30 degrees above the horizontal, we can use the formula for projectile motion.
First, we need to find the horizontal and vertical components of the initial velocity.
The horizontal component (vx) is given by vx = v * cos(θ), and the vertical component (vy) is vy = v * sin(θ). Using these, we can determine the time spent in the air and then the horizontal distance traveled.
For the horizontal component: vx = 25 m/s * cos(30°) = 21.65 m/s.
The time of flight is determined by the vertical motion.
The initial vertical velocity is vy = 25 m/s * sin(30°) = 12.5 m/s.
The time taken to reach the maximum height (tup) can be found using tup = vy / g, where g is the acceleration due to gravity (9.81 m/s2).
The total time in the air (T) will be twice this time because it has to go up and come back down.
Calculating the time: tup = 12.5 m/s / 9.81 m/s2 = 1.27 s and T = 2 * tup = 2.54 s.
Finally, the horizontal distance (d) can be calculated using d = vx * T = 21.65 m/s * 2.54 s = 55 m (rounded to two decimal places).
None of the given options are correct.
Question: A projectile is launched with an initial velocity of 25 m/s at an angle of 30 degrees above the horizontal. Ignoring air resistance, calculate the horizontal distance traveled by the projectile before hitting the ground. Choose the correct answer from the options below:
a. 20.6 meters
b. 34.1 meters
c. 42.3 meters
d. 48.9 meters