Question

Sam kicks a soccer ball with an initial velocity of 40 ms-1. If he kicked the ball at an angle of 550, how far did he kick the ball? What was its maximum height? How long was it in the air.

Answer 1

1. Range = 153.42 m.

2. Maximum height = 54.78 m

3. Time of flight = 6.69 s

Step-by-step 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 follow:

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