Question

A soccer ball is kicked at an angle of 59 ◦ to the horizontal with an

initial speed of 20 m/s. Assume for the moment that we can neglect air

resistance. (a) For how much time is the ball in the air? (b) How far does it go

(horizontal distance along the field)? (c) How high does it go?

Short answer

(b) The horizontal distance a soccer ball will go is,$R=36.00\mathrm{m}$

Step-by-step solution

Identification of given data

1. Projectile angle is,$\theta ={59}^{°}$.

2. The ball’s initial speed is given as,$\upsilon =$20m/s.

Concept of projectile motion

An object or body, when it is kicked/ thrown and moves under the influence of

gravity, this object is known as a projectile, and the motion of this projectile is

called projectile motion. The angle at which it is aimed at a target is known as

the projectile angle.

b) Determination of range of soccer ball

The range of the soccer ball can be evaluated as follows,

$R=\frac{{\upsilon }^{2\sin 2\theta }}{g}$

Substitute the values in the above equation, and we get,

$$\begin{gathered} R=\frac{{\left(20\mathrm{m}/\mathrm{s}\right)}^2\sin 2\left(59°\right)}{9.81\mathrm{m}/{\mathrm{s}}^2} \\ R=36.00\mathrm{m} \end{gathered}$$

Thus, the maximum horizontal distance covered is 36.00m .