Final answer:
The maximum height of a projectile fired with an initial speed of 15 m/s at an angle of 68 degrees can be calculated using kinematic equations and is found to be approximately 9.8 meters.
Step-by-step explanation:
Calculating the Maximum Height of a Projectile
To determine the maximum height of a projectile's trajectory, we can apply the kinematic equations for projectile motion. Given an initial speed (v) of 15 m/s and a projection angle (θ) of 68 degrees with respect to the horizontal, we first need to calculate the initial vertical component of the velocity (vy).
The initial vertical velocity component is given by:
vy = v × sin(θ)
Using the given values:
vy = 15 m/s × sin(68°) ≈ 15 m/s × 0.927 ≈ 13.9 m/s
At the projectile's maximum height, the vertical velocity will be 0 m/s. Using the kinematic formula:
v² = u² + 2as,
where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement. Since we are looking for the maximum height, v = 0 m/s, u = 13.9 m/s, a = -9.8 m/s² (the acceleration due to gravity, acting downwards).
Rearranging the formula to solve for s (maximum height) gives us:
s = –(u²) / (2a)
Plugging in the known values:
s = –(13.9 m/s)² / (2 × -9.8 m/s²) ≈ 9.8 m
Therefore, the maximum height of the projectile's trajectory is approximately 9.8 meters, which corresponds to option c).