Respuesta :
A football player kicks a football off a tee with a speed of 16 m/s at an angle of 63°
The horizontal and vertical speed of the ball is given by
[tex]\begin{gathered} v_x=v\cos (\theta) \\ v_y=v\sin (\theta) \end{gathered}[/tex]Where v = 16 m/s and θ = 63°
[tex]\begin{gathered} v_x=16\cos (63\degree)=7.26\; \frac{m}{s} \\ v_y=16\sin (63\degree)=14.26\; \frac{m}{s} \end{gathered}[/tex]How far is the ball from the football player when it lands?
The range of the ball is given by
[tex]x=v_x\times t[/tex]Where t is the time the ball remains in the air.
The time (t) can be found as
[tex]y=v_yt+\frac{1}{2}at^2[/tex]y = 0 when the ball is in the air.
The acceleration is due to gravity (-9.8 m/s²)
[tex]\begin{gathered} 0=14.26t+\frac{1}{2}(-9.8)t^2 \\ 0=14.26t-4.9t^2 \\ 0=t(14.26-4.9t) \\ 0=14.26-4.9t \\ 4.9t=14.26 \\ t=\frac{14.26}{4.9} \\ t=2.91\; s \end{gathered}[/tex]Finally, the range is
[tex]x=v_x\times t=7.26\times2.91=21.13\; m[/tex]Therefore, the ball will land 21.13 meters far from the football player.
How much farther would the ball go if he kicked it with the same speed, but at a 45° angle?
We need to repeat the above calculations
The horizontal and vertical speed of the ball is given by
[tex]\begin{gathered} v_x=v\cos (\theta)=16\cos (45\degree)=11.31\; \frac{m}{s} \\ v_y=v\sin (\theta)=16\sin (45\degree)=11.31\; \frac{m}{s} \end{gathered}[/tex]The time (t) is given by
[tex]\begin{gathered} y=v_yt+\frac{1}{2}at^2 \\ 0=11.31_{}t+\frac{1}{2}(-9.8)t^2 \\ 0=11.31_{}t-4.9t^2 \\ 0=11.31_{}-4.9t \\ 4.9t=11.31_{} \\ t=\frac{11.31_{}}{4.9} \\ t=2.31\; s \end{gathered}[/tex]Finally, the range is
[tex]x=v_x\times t=11.31\times2.31=26.13\; m[/tex]Therefore, the ball will land 26.13 meters far from the football player.
Which ball will land first: the ball kicked at 16 m/s and at a 63° angle, or one kicked at 9 m/s and at a 45° angle?
The ball kicked at 16 m/s and at a 63° angle takes 2.91 s to land.
The ball kicked at 9 m/s and at a 45° angle will take
[tex]v_y=9\sin (45\degree)=6.36\; \frac{m}{s}[/tex][tex]\begin{gathered} y=v_yt+\frac{1}{2}at^2 \\ 0=6.36t+\frac{1}{2}(-9.8)t^2 \\ 0=6.36t-4.9t^2 \\ 0=6.36-4.9t \\ 4.9t=6.36 \\ t=\frac{6.36}{4.9} \\ t=1.30\; s \end{gathered}[/tex]So, the ball kicked at 9 m/s and at a 45° angle takes 1.30 s to land.
Therefore, the ball kicked at 9 m/s and at a 45° angle will land first.
