A soccer ball is kicked from the ground with an initial speed v at an upward angle θ. A player a distance d away in the direction of the kick starts running towards the ball at that instant. What must be his average speed if he is to meet the ball just before it hits the ground? Give your answer in terms of g and the given variables (neglecting air resistance).

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mustafaazeem664 mustafaazeem664 · Answer 1 · 2020-02-19T08:40:15+01:00

Answer:

The speed of player is given by

[tex]V=\frac{2v^{2} 2sin\alpha.cos\alpha-gd }{2vsin\alpha}[/tex]

Explanation:

The time of flight for a projectile motion is given by

[tex]T=\frac{2vsin\alpha }{g}[/tex] (i)

where t is the time of flight, v is the initial speed, and α is the angle.

Now the person must also reach the impact point of ball in the same time as above.

Now the total distance D the player needs to cover is basically R horizontal range of projectile minus the distance d, range R is given by,

[tex]R=\frac{2v^{2} sin2\alpha }{g}[/tex]

Now the distance the player must cover is given by

D= R-d

D= [tex]\frac{2v^{2} sin2\alpha }{g}[/tex] - d

[tex]D=\frac{2v^{2}sin2\alpha-gd}{g}[/tex] (ii)

Now the average speed of player is given by

[tex]V=\frac{D}{T}[/tex] (iii)

Replacing the values of D and T from eq. (i) and (ii) in eq. (iii).

[tex]V=\frac{\frac{2v^{2} sin2\alpha-gd }{g}}{\frac{2vsin\alpha }{g} }[/tex]

[tex]V=\frac{2v^{2} 2sin\alpha.cos\alpha-gd }{2vsin\alpha}[/tex]