Question

A soccer ball on the ground was kicked with an initial velocity of 35 ft/sec at an elevation angle of 40º. Create a set of parametric equations to model the path of the soccer ball. Approximately how far away did the soccer ball hit the ground?
19 feet

26 feet

38 feet

52 feet

Answer 1

Answer:

38 feet.

Step-by-step explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 35 ft/s

Angle of projection (θ) = 40º

Acceleration due to gravity (g) = 9.8 m/s²

Range (R) =?

Next, we shall convert 9.8 m/s² to ft/s². This can be obtained as follow:

0.3048 m/s² = 1 ft/s²

Therefore,

9.8 m/s² = 9.8 m/s² × 1 ft/s²/ 0.3048 m/s²

9.8 m/s² = 32.15 ft/s²

Thus, 9.8 m/s² is equivalent to 32.15 ft/s².

Finally, we shall determine how far away the soccer ball hit the ground (i.e range) as shown below:

Initial velocity (u) = 35 ft/s

Angle of projection (θ) = 40º

Acceleration due to gravity (g) = 32.15 ft/s²

Range (R) =?

R = u² × Sine 2θ / g

R = 35² × Sine (2×40) / 32.15

R = 1225 × Sine 80/ 32.15

R = 1225 × 0.9848 / 32.15

R = 37.5 ≈ 38 feet.

Thus, the soccer ball hit the ground 38 feet away.

Answer 2

Answer:

The anwser is C.

Step-by-step explanation: