Answer:
1) The ball will land approximately 141.4 m away from where it was kicked
2) The maximum height reached is approximately 61.22 m
3) The total hang time is approximately 7.07 seconds
Explanation:
First we list out the known parameters
The direction in which the punt is kicked, θ = 60°
The speed with which the punk is kicked, v₀ = 40 m/s
1) The distance away (the range, R) the ball will land is given by the equation;
[tex]R = \dfrac{v_0^2 \times sin(2\cdot \theta)}{g}[/tex]
Where;
g = The acceleration due to gravity = 9.8 m/s²
Substituting the values, gives;
[tex]R = \dfrac{40^2 \times sin(2\times60^{\circ })}{9.8} \approx 141.4 \ m[/tex]
The distance away the ball will land ≈ 141.4 m
2) The maximum height is given m=by the following equation for vertical motion;
(v₀ × sin(θ))² = 2 × g × [tex]h_{max}[/tex]
Obtained from the equation of motion, v² = u² - 2×g×h where at maximum height;
v = 0 at maximum height, the ball stops upward motion
h = Height of motion = [tex]h_{max}[/tex] at maximum height
v, and u are the vertical component of the velocity, v₀
Substituting the values in the equation for the maximum height, gives;
(40 × sin(60))² = 2 × 9.8 × [tex]h_{max}[/tex]
(40 × (√3)/2)² = 1200 = 19.6 × [tex]h_{max}[/tex]
1200 = 19.6 × [tex]h_{max}[/tex]
[tex]h_{max}[/tex] = 1200/19.6 ≈ 61.22 m
The maximum height reached ≈ 61.22 m
3) The time to maximum height is given by the following equation of motion;
[tex]v_y[/tex] = v₀ × sin(θ) - g × t
[tex]v_y[/tex] = 0 at maximum height
We therefore have;
v₀ × sin(θ) = g × t
t = v₀ × sin(θ)/g = 40 × sin(60°)/9.8 = (40 ×(√3)/2)/9.8 = 3.535 seconds
Given that the motion of the ball is symmetrical about the maximum height, we have;
The time from ground level to maximum height = The time from the maximum height back to the ground level
The total time hang time = The time from ground level to maximum height + The time from the maximum height back to the ground level
The total time hang time ≈ 3.535 + 3.535 ≈ 7.07 seconds.
The total time hang time ≈ 7.07 seconds.
