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.
