Final answer:
The maximum horizontal distance that the boy can throw the stone up to is approximately 10√2 meters.
Therefore, the correct answer is: option C). 10√2m
Step-by-step explanation:
To find the maximum horizontal distance the boy can throw the stone, we need to analyze the projectile motion.
The horizontal distance traveled by a projectile depends on its initial velocity and the time of flight. Since the stone is thrown horizontally at its maximum height, the initial vertical velocity is zero.
Using the equation for projectile motion: d = vx * t { where d is the horizontal distance, vx is the initial horizontal velocity, and t is the time of flight.}
Since the initial vertical velocity is zero, the initial horizontal velocity is equal to the maximum horizontal velocity the boy can throw the stone.
Therefore, the maximum horizontal distance is: d = (v0 * cos(θ)) * t { where v0 is the initial velocity of the stone, cos(θ) is the cosine of the launch angle, and t is the time of flight}.
Since we are given the maximum height of 10 m, there is no mention of the launch angle. Therefore, we can assume the stone is thrown vertically upward with an initial velocity of 10 m/s.
Using the equation: t = √(2 * h / g) { where h is the maximum height and g is the acceleration due to gravity (approximately 9.8 m/s²), we can find the time of flight. }
Plugging the values into the equations:
t = √(2 * 10 / 9.8) = √(20 / 9.8) ≈ 1.44 s
Finally, substituting the values into the formula for the maximum horizontal distance:
d = (10 * cos(0)) * 1.44
= 10 * 1 * √2
= 10√2
Therefore, the maximum horizontal distance that the boy can throw the stone up to is approximately 10√2 meters, option C.