1) a soccer ball is kicked across a field at an angle of 30° with an initial speed of 16 m/s. write a set of…

1) a soccer ball is kicked across a field at an angle of 30° with an initial speed of 16 m/s. write a set of parametric equations for the motion of the soccer ball.\n2) a golf ball is struck across a flat fairway at an angle of 45° with an initial speed of 136 ft/s. write a set of parametric equations for the motion of the golf ball.\n3) a snowball is thrown at an angle of 60° with an initial speed of 45 ft/s and an initial height of 5 ft. write a set of parametric equations for the motion of the snowball.\n4) a cannon ball is fired across a flat field at an angle of 45° with an initial speed of 103 ft/s and an initial height of 31 ft.\ni) write a set of parametric equations for the motion of the cannon ball.\nii) determine how long the cannon ball was in the air.\niii) determine how far the cannon ball traveled in the air.\niv) determine when the cannon ball reached its maximum height.\nv) determine the maximum height reached by the cannon ball.\nvi) sketch the graph of the parametric equations.

1) a soccer ball is kicked across a field at an angle of 30° with an initial speed of 16 m/s. write a set of parametric equations for the motion of the soccer ball.\n2) a golf ball is struck across a flat fairway at an angle of 45° with an initial speed of 136 ft/s. write a set of parametric equations for the motion of the golf ball.\n3) a snowball is thrown at an angle of 60° with an initial speed of 45 ft/s and an initial height of 5 ft. write a set of parametric equations for the motion of the snowball.\n4) a cannon ball is fired across a flat field at an angle of 45° with an initial speed of 103 ft/s and an initial height of 31 ft.\ni) write a set of parametric equations for the motion of the cannon ball.\nii) determine how long the cannon ball was in the air.\niii) determine how far the cannon ball traveled in the air.\niv) determine when the cannon ball reached its maximum height.\nv) determine the maximum height reached by the cannon ball.\nvi) sketch the graph of the parametric equations.

Answer

Explanation:

Step1: Parametric equations for projectile motion

The general parametric equations for projectile motion are (x = v_{0}\cos\theta t) and (y=v_{0}\sin\theta t-\frac{1}{2}gt^{2}+y_{0}), where (v_{0}) is the initial speed, (\theta) is the launch angle, (g) is the acceleration due to gravity ((g = 9.8\ m/s^{2}) or (g = 32\ ft/s^{2})), and (y_{0}) is the initial height.

Step2: For the soccer - ball problem (Problem 1)

Given (v_{0}=16\ m/s), (\theta = 30^{\circ}), and (y_{0} = 0) (assuming it is kicked from ground - level). (\cos30^{\circ}=\frac{\sqrt{3}}{2}), (\sin30^{\circ}=\frac{1}{2}), (g = 9.8\ m/s^{2}) (x=(16\cos30^{\circ})t=16\times\frac{\sqrt{3}}{2}t = 8\sqrt{3}t) (y=(16\sin30^{\circ})t-\frac{1}{2}(9.8)t^{2}=8t - 4.9t^{2})

Step3: For the golf - ball problem (Problem 2)

Given (v_{0}=136\ ft/s), (\theta = 45^{\circ}), (y_{0}=0) (flat fairway), (g = 32\ ft/s^{2}) (\cos45^{\circ}=\sin45^{\circ}=\frac{\sqrt{2}}{2}) (x=(136\cos45^{\circ})t=136\times\frac{\sqrt{2}}{2}t = 68\sqrt{2}t) (y=(136\sin45^{\circ})t-\frac{1}{2}(32)t^{2}=68\sqrt{2}t-16t^{2})

Step4: For the snow - ball problem (Problem 3)

Given (v_{0}=45\ ft/s), (\theta = 60^{\circ}), (y_{0}=5\ ft), (g = 32\ ft/s^{2}) (\cos60^{\circ}=\frac{1}{2}), (\sin60^{\circ}=\frac{\sqrt{3}}{2}) (x=(45\cos60^{\circ})t=\frac{45}{2}t) (y=(45\sin60^{\circ})t-\frac{1}{2}(32)t^{2}+5=\frac{45\sqrt{3}}{2}t-16t^{2}+5)

Step5: For the cannon - ball problem (Problem 4i)

Given (v_{0}=103\ ft/s), (\theta = 45^{\circ}), (y_{0}=31\ ft), (g = 32\ ft/s^{2}) (\cos45^{\circ}=\sin45^{\circ}=\frac{\sqrt{2}}{2}) (x=(103\cos45^{\circ})t=\frac{103\sqrt{2}}{2}t) (y=(103\sin45^{\circ})t-\frac{1}{2}(32)t^{2}+31=\frac{103\sqrt{2}}{2}t-16t^{2}+31)

Step6: For the cannon - ball problem (Problem 4ii)

Set (y = 0) in (y=\frac{103\sqrt{2}}{2}t-16t^{2}+31) (16t^{2}-\frac{103\sqrt{2}}{2}t - 31=0) Using the quadratic formula (t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}) where (a = 16), (b=-\frac{103\sqrt{2}}{2}), (c=-31) (t=\frac{\frac{103\sqrt{2}}{2}\pm\sqrt{(\frac{103\sqrt{2}}{2})^{2}-4\times16\times(-31)}}{2\times16}) (t=\frac{\frac{103\sqrt{2}}{2}\pm\sqrt{\frac{103^{2}\times2}{4}+1984}}{32}) (t=\frac{\frac{103\sqrt{2}}{2}\pm\sqrt{\frac{21218}{4}+1984}}{32}=\frac{\frac{103\sqrt{2}}{2}\pm\sqrt{\frac{21218 + 7936}{4}}}{32}=\frac{\frac{103\sqrt{2}}{2}\pm\sqrt{\frac{29154}{4}}}{32}) (t=\frac{\frac{103\sqrt{2}}{2}\pm\frac{\sqrt{29154}}{2}}{32}) We take the positive root (t=\frac{\frac{103\sqrt{2}+\sqrt{29154}}{2}}{32}\approx\frac{\frac{103\times1.414+170.74}{2}}{32}=\frac{\frac{145.642 + 170.74}{2}}{32}=\frac{\frac{316.382}{2}}{32}=\frac{158.191}{32}\approx4.94\ s)

Step7: For the cannon - ball problem (Problem 4iii)

Substitute (t\approx4.94) into (x=\frac{103\sqrt{2}}{2}t) (x=\frac{103\sqrt{2}}{2}\times4.94\approx\frac{103\times1.414}{2}\times4.94\approx(73.14)\times4.94\approx361.3\ ft)

Step8: For the cannon - ball problem (Problem 4iv)

The time to reach maximum height is given by (t =-\frac{b}{2a}) in (y = at^{2}+bt + c) For (y=-16t^{2}+\frac{103\sqrt{2}}{2}t + 31), (a=-16), (b = \frac{103\sqrt{2}}{2}) (t=\frac{-\frac{103\sqrt{2}}{2}}{2\times(-16)}=\frac{103\sqrt{2}}{64}\approx\frac{103\times1.414}{64}\approx\frac{145.642}{64}\approx2.28\ s)

Step9: For the cannon - ball problem (Problem 4v)

Substitute (t=\frac{103\sqrt{2}}{64}) into (y=-16t^{2}+\frac{103\sqrt{2}}{2}t + 31) (y=-16\times(\frac{103\sqrt{2}}{64})^{2}+\frac{103\sqrt{2}}{2}\times\frac{103\sqrt{2}}{64}+31) (y=-16\times\frac{103^{2}\times2}{64^{2}}+\frac{103^{2}\times2}{128}+31) (y=-\frac{103^{2}}{128}+\frac{103^{2}}{64}+31=\frac{103^{2}}{128}+31=\frac{10609}{128}+31\approx82.9+31 = 113.9\ ft)

Answer:

  1. (x = 8\sqrt{3}t), (y = 8t-4.9t^{2})
  2. (x = 68\sqrt{2}t), (y = 68\sqrt{2}t - 16t^{2})
  3. (x=\frac{45}{2}t), (y=\frac{45\sqrt{3}}{2}t-16t^{2}+5) 4i. (x=\frac{103\sqrt{2}}{2}t), (y=\frac{103\sqrt{2}}{2}t-16t^{2}+31) 4ii. (t\approx4.94\ s) 4iii. (x\approx361.3\ ft) 4iv. (t\approx2.28\ s) 4v. (y\approx113.9\ ft)