Question #150730
Winn and James are playing football in the field. Winn kicked the football so that it leaves the ground with an initial velocity of 23 m/s with an angle of projection of 20˚.
(a) What are the horizonal and vertical components of the ball’s initial velocity?
(b) At which time will the ball reach the highest peak of the trajectory?
(c) How long will the ball stay in flight? What is the total time of flight?
(d) What is the ball’s maximum height during its flight?
1
Expert's answer
2020-12-14T12:12:48-0500

Let's write the equations of motion of the football in horizontal and vertical directions:


x=v0tcosθ,(1)x=v_0tcos\theta, (1)y=v0tsinθ12gt2,(2)y=v_0tsin\theta-\dfrac{1}{2}gt^2, (2)

here, xx is the horizontal displacement of the football, v0=23 msv_0=23\ \dfrac{m}{s} is the initial velocity of the football, tt is the total time of flight of the football, θ=20\theta=20^{\circ} is the launch angle, yy is the vertical displacement of the football (or the height) and g=9.8 ms2g=9.8\ \dfrac{m}{s^2} is the acceleration due to gravity.

a) Let's find the horizontal and vertical components of the football's initial velocity:


v0x=v0cosθ=23 mscos20=21.6 ms,v_{0x}=v_0cos\theta=23\ \dfrac{m}{s}\cdot cos20^{\circ}=21.6\ \dfrac{m}{s},v0y=v0sinθ=23 mssin20=7.87 ms.v_{0y}=v_0sin\theta=23\ \dfrac{m}{s}\cdot sin20^{\circ}=7.87\ \dfrac{m}{s}.

b) Let's first find the time that the football takes to reach the highest peak of the trajectory:


vy=v0sinθgtrise,v_y=v_0sin\theta-gt_{rise},0=v0sinθgtrise,0=v_0sin\theta-gt_{rise},trise=v0sinθg=23 mssin209.8 ms2=0.8 s.t_{rise}=\dfrac{v_0sin\theta}{g}=\dfrac{23\ \dfrac{m}{s}\cdot sin20^{\circ}}{9.8\ \dfrac{m}{s^2}}=0.8\ s.


c) Then, we can find the total time of flight of the footbal:


t=2trise=20.8 s=1.6 s.t=2t_{rise}=2\cdot 0.8\ s=1.6\ s.

d) Finally, we can substitute triset_{rise} into the second equation and find the maximum height:


ymax=v0sinθv0sinθg12g(v0sinθg)2,y_{max}=v_0sin\theta\cdot\dfrac{v_0sin\theta}{g}-\dfrac{1}{2}g(\dfrac{v_0sin\theta}{g})^2,ymax=v02sin2θ2g,y_{max}=\dfrac{v_0^2sin^2\theta}{2g},ymax=(23 ms)2sin22029.8 ms2=3.16 m.y_{max}=\dfrac{(23\ \dfrac{m}{s})^2\cdot sin^220^{\circ}}{2\cdot 9.8\ \dfrac{m}{s^2}}=3.16\ m.

Answer:

a) v0x=21.6 ms,v0y=7.87 ms.v_{0x}=21.6\ \dfrac{m}{s}, v_{0y}=7.87\ \dfrac{m}{s}.

b) trise=0.8 s.t_{rise}=0.8\ s.

c) t=1.6 s.t=1.6\ s.

d) ymax=3.16 m.y_{max}=3.16\ m.


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