Can a goalkeeper at her/ his goal kick a soccer ball into the opponent’s goal without the ball touching the ground? The distance will be about 95m. A goalkeeper can give the ball a speed of 30m/s.

Short Answer

Expert verified

No goalkeeper cannot goal kick a soccer ball into the opponent’s goal without the ball touching the ground

Step by step solution

01

Given data

The problem is asking us to calculate the range. If the initial velocity of the ball is 30 m/s. To maximum the range of the ball the angle should be kicked at 45 degree.

The formula of range is

Rx=Vi2sin2θig

02

Range of the ball

Putting the value of range in the equation

Rx=Vi2sin2θigIfθ=45Rx=302sin909.8Rx=91.8m

The range of the ball is 91.8 meter, which is less 95 meter. No goalkeeper cannot goal kick a soccer ball into the opponent’s goal without the ball touching the ground.

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Most popular questions from this chapter

Answer the following questions for projectile motion on level ground assuming negligible air resistance (the initial angle being neither \(0^\circ \) nor \(90^\circ \)):

(a) Is the velocity ever zero?

(b) When is the velocity a minimum? A maximum?

(c) Can the velocity ever be the same as the initial velocity at a time other than at \(t = 0\)?

(d) Can the speed ever be the same as the initial speed at a time other than at \(t = 0\)?

(a) Repeat the problem two problems prior, but for the second leg you walk \(20.0{\rm{ m}}\) in a direction \(40.0^\circ \) north of east (which is equivalent to subtracting \({\rm{B}}\) from \({\rm{A}}\) —that is, to finding \({\rm{R'}} = {\rm{A}} - {\rm{B}}\)).

(b) Repeat the problem two problems prior, but now you first walk \(20.0{\rm{ m}}\) in a direction \(40.0^\circ \) south of west and then \(12.0{\rm{ m}}\) in a direction \(20.0^\circ \) east of south (which is equivalent to subtracting \({\rm{A}}\) from \({\rm{B}}\) —that is, to finding \({\rm{R''}} = {\rm{B}} - {\rm{A}} = - {\rm{R'}}\)). Show that this is the case.

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