A small ball is launched at an angle of 30.0 degrees about the horizontal. It reaches a maximum height of 2.5m with a respect to the launch position.
What is the initial velocity of the ball when its launched?
What is the range, defined as a horizontal distance traveled until it returns to his original height. (you can ignore air resistance)
2.5 m/s
To find the initial velocity of the ball when it is launched, we can use the fact that the maximum height is reached when the vertical component of the velocity is zero.
We can use the following kinematic equation to find the vertical component of the velocity at the maximum height:
Vf^2 = Vi^2 + 2 * a * Δy,
where Vf is the final velocity (which is zero at maximum height), Vi is the initial velocity, a is the acceleration due to gravity (-9.8 m/s^2), and Δy is the vertical displacement (2.5 m).
Plugging the values into the equation:
0 = Vi^2 + 2 * (-9.8 m/s^2) * 2.5 m,
Simplifying the equation:
0 = Vi^2 - 49 m^2/s^2.
Solving for Vi, we find:
Vi = √(49 m^2/s^2) = 7 m/s.
Therefore, the initial velocity of the ball when it is launched is 7 m/s.
To find the range of the ball, we can use the following equation:
Range = (Vi^2 * sin(2θ)) / g,
where Vi is the initial velocity (7 m/s), θ is the launch angle (30 degrees), and g is the acceleration due to gravity (9.8 m/s^2).
Plugging the values into the equation:
Range = (7 m/s)^2 * sin(2 * 30 degrees) / 9.8 m/s^2,
Simplifying the equation:
Range = (49 m^2/s^2) * sin(60 degrees) / 9.8 m/s^2,
Range = (49 m^2/s^2) * √(3)/2 / 9.8 m/s^2,
Range = 24.5 m * √(3) / 9.8,
Range ≈ 14.1 m.
Therefore, the range of the ball, defined as the horizontal distance traveled until it returns to its original height, is approximately 14.1 meters.