Sam kicks a soccer ball with an initial velocity of 40 ms-1. If he kicked the ball at an angle of 55°, how far did he kick the ball? What was its maximum height? How long was it in the air?

Question

Calista · Answer

1. Range = 153.42 m. 2. Maximum height = 54.78 m 3. Time of flight = 6.69 s  Explanation: From the question given above, the following data were obtained: Initial velocity (u) = 40 ms¯¹ Angle of projection (θ) = 55°  1. Determination of the range: Initial velocity (u) = 40 ms¯¹ Angle of projection (θ) = 55° Acceleration due to gravity (g) = 9.8 ms¯² Range (R) =? The range i.e how far the ball went can be obtained as follows: R = u² Sine 2θ / g R = 40² × Sine (2×55) / 9.8 R = 1600 × Sine 110 / 9.8 R = 1600 × 0.9397 /9.8 R = 153.42 m  2. Determination of the maximum height: Initial velocity (u) = 40 ms¯¹ Angle of projection (θ) = 55° Acceleration due to gravity (g) = 9.8 ms¯² Maximum height (H) =? H = u² Sine² θ / 2g H =  40² × (Sine 55)² / 2 × 9.8 H =  1600 × (0.8192)² / 19.6 H = 54.78 m  3. Determination of the time of flight: Initial velocity (u) = 40 ms¯¹ Angle of projection (θ) = 55° Acceleration due to gravity (g) = 9.8 ms¯² Time of flight (T) =? T = 2u Sine θ / g T = 2 × 40 × Sine 55 / 9.8 T = 80 × 0.8192 / 9.8 T = 6.69 s

Calculating the Range, Maximum Height, and Time of Flight of a Soccer Ball

Solving for the Range:

Finding the Maximum Height:

Determining the Time of Flight:

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