Abbie kicks a ball from one end of the 100m field to the other if the ball is in the air for 4 seconds determine the angle the ball was kicked at and the initial velocity. Solve the question using kinematics formulas.
To solve this question, we can use the kinematics formulas for projectile motion. The two equations we'll be using are:
1) Horizontal displacement (Δx) = initial velocity (V₀) * time (t) * cos(θ)
2) Vertical displacement (Δy) = initial velocity (V₀) * time (t) * sin(θ) - (1/2) * acceleration due to gravity (g) * t²
Given that the field length is 100m and the ball is in the air for 4 seconds, we know that the horizontal displacement (Δx) is 100m and the time (t) is 4 seconds.
Now, let's solve for the initial velocity (V₀) and the angle (θ):
1) Δx = V₀ * t * cos(θ)
100m = V₀ * 4s * cos(θ)
Divide both sides by (4s * cos(θ)):
25m = V₀ * cos(θ)
2) Δy = V₀ * t * sin(θ) - (1/2) * g * t²
Assuming the ball starts at ground level, the vertical displacement (Δy) is 0. Therefore:
0 = V₀ * 4s * sin(θ) - (1/2) * 9.8 m/s² * (4s)²
0 = V₀ * 4s * sin(θ) - 78.4m
Now, we have two equations with two unknowns (V₀ and θ). We can use substitution to solve for both.
From equation 1, we have:
25m = V₀ * cos(θ) --> (Equation A)
From equation 2, we have:
0 = V₀ * 4s * sin(θ) - 78.4m --> (Equation B)
Now, let's solve equation A for V₀:
V₀ = 25m / cos(θ)
Substitute this value of V₀ into equation B:
0 = (25m / cos(θ)) * 4s * sin(θ) - 78.4m
0 = (100m * sin(θ)) - 78.4m
Rearrange the equation:
78.4m = 100m * sin(θ)
sin(θ) = 78.4m / 100m
sin(θ) = 0.784
Taking the arcsin of both sides:
θ = arcsin(0.784)
θ ≈ 51.81° (rounded to 2 decimal places)
Now, substitute this value of θ into equation A to find V₀:
25m = V₀ * cos(51.81°)
V₀ = 25m / cos(51.81°)
V₀ ≈ 40.10 m/s (rounded to 2 decimal places)
Therefore, the ball was kicked at an angle of approximately 51.81° and the initial velocity was approximately 40.10 m/s.