Final answer:
The range of the ball is approximately 277.13 m and the greatest height realized is 80 m.
Step-by-step explanation:
To find the range and greatest height of the ball, we can use the equations of projectile motion.
First, let's find the time it takes for the ball to reach the highest point of its trajectory. The vertical component of the initial velocity is given by:
Vy = V * sin(theta)
where V is the initial velocity (80 m/s) and theta is the launch angle (30 degrees). So, Vy = 80 * sin(30) = 40 m/s. At the highest point, the vertical component of the velocity will be zero, so we can use the following equation:
Vy = Vy0 - g * t
where Vy0 is the initial vertical component of the velocity (40 m/s), g is the acceleration due to gravity (10 m/s^2), and t is the time it takes to reach the highest point. Solving for t:
0 = 40 - 10 * t
t = 4 s
Next, let's find the maximum height reached by the ball. We can use the following equation:
H = Vy0 * t - 0.5 * g * t^2
where H is the maximum height. Substituting the values:
H = 40 * 4 - 0.5 * 10 * 4^2
H = 160 - 80
H = 80 m
Finally, let's find the range of the ball. We can use the equation:
R = Vx * t
where Vx is the horizontal component of the initial velocity and t is the time taken to reach the highest point (4 s). Since there is no acceleration in the horizontal direction, Vx remains constant throughout the motion. So:
R = V * cos(theta) * t
R = 80 * cos(30) * 4
R = 80 * √3/2 * 4
R = 80 * 2√3
R = 160√3 m
Therefore, the range of the ball is approximately 277.13 m and the greatest height realized is 80 m.