How to Calculate the Distance, Speed, and Height of a Soccer Ball in Motion
How can we calculate the distance, speed, and height of a soccer ball in motion?
1) What is the initial speed of the ball?
2) What is the initial angle 0 of the ball with respect to the ground?
3) What is the maximum height the ball goes above the ground?
4) How far from where it was kicked will the ball land?
5) What is the speed of the ball 2.5 seconds after it was kicked?
6) How high above the ground is the ball 2.5 seconds after it is kicked?
Answers:
4) The distance from where the ball was kicked is 38.06 meters.
5) The speed of the ball 2.5 seconds after it was kicked is 13.82 m/s.
6) The ball is 21.88 meters above the ground 2.5 seconds after it is kicked.
4) To calculate the distance from where the ball was kicked, we need to find the time it takes to reach the ground. We can use the fact that the vertical displacement of the ball is zero at the highest point. Using the formula vf = vi + at, the time it takes to reach maximum height is t = vf / g where g is the acceleration due to gravity which is -9.8 m/s² since it is downward and vf is the final velocity which is 0 because the ball comes to rest at the highest point. t = 17 / 9.8 = 1.73 s. This means the total time for the ball to hit the ground is 2 x 1.73 = 3.46 s. Using the formula for horizontal distance traveled d = vt, we get d = 11 x 3.46 = 38.06 m. So, the distance from where the ball was kicked will be 38.06 meters.
5) To calculate the speed of the ball 2.5 seconds after it was kicked, we need to find the horizontal and vertical components of the velocity of the ball at 2.5 seconds. The horizontal component is constant, so it will still be 11 m/s. To find the vertical component, we use the formula vf = vi + at where vi is initial velocity, a is acceleration due to gravity which is -9.8 m/s² and t is the time which is 2.5 seconds. vf = 17 + (-9.8 x 2.5) = -7.5 m/s. Since the ball is moving downward, the velocity is negative. Therefore, the speed of the ball 2.5 seconds after it was kicked is sqrt(11² + (-7.5)²) = 13.82 m/s.
6) To calculate how high above the ground is the ball 2.5 seconds after it is kicked, we use the formula for the displacement of an object in the vertical direction y = vi*t + (1/2)*a*t² where vi is initial velocity, a is acceleration due to gravity which is -9.8 m/s² and t is the time which is 2.5 seconds. y = 17*2.5 + (1/2)*(-9.8)*(2.5)² = 21.88 m. So, the ball is 21.88 m above the ground 2.5 seconds after it is kicked.