A ball is kicked so that it has an initial velocity of 12 m s –1 at an angle of 50 to horizontal ground.Assume that air resistance is negligible.5012 m s–1What is the maximum height above the ground that is reached by the ball?A 0.47 m B 3.0 m C 4.3 m D 7.3 m
Question
A ball is kicked so that it has an initial velocity of 12 m s –1 at an angle of 50 to horizontal ground.Assume that air resistance is negligible.5012 m s–1What is the maximum height above the ground that is reached by the ball?A 0.47 m B 3.0 m C 4.3 m D 7.3 m
Solution
To solve this problem, we need to use the equations of motion. The maximum height (H) can be found using the formula:
H = (u^2 * sin^2(θ)) / (2 * g)
where:
- u is the initial velocity (12 m/s),
- θ is the angle of projection (50 degrees),
- g is the acceleration due to gravity (approximately 9.8 m/s^2).
First, we need to convert the angle from degrees to radians because the trigonometric functions in most calculators use radians.
50 degrees = 50 * π / 180 = 0.8727 radians
Then, we substitute the values into the formula:
H = (12^2 * sin^2(0.8727)) / (2 * 9.8) H = (144 * 0.766) / 19.6 H = 110.304 / 19.6 H = 5.63 m
So, the maximum height reached by the ball is approximately 5.63 meters. However, this option is not available in the choices given. Please check the values or the choices again.
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