Answer:
The horizontal range is maximum when θ is equal to 45°
Explanation:
Projectile Motion
It's known as the type of motion that experiences an object that is projected near the Earth's surface and moves along a curved path exclusively under the action of gravity.
Being vo the initial speed of the object, θ the initial launch angle, and [tex]g=9.8\ m/s^2[/tex] the acceleration of gravity, then the maximum horizontal distance traveled by the object (also called Range) is:
[tex]\displaystyle d={\frac {v_o^{2}\sin(2\theta )}{g}}[/tex]
Since the sine of an angle is bounded between -1 and 1, the maximum range will occur when the sine is 1:
[tex]\sin(2\theta)=1[/tex]
The only angle between 0° and 360° whose sine is 1 is 90°, thus:
[tex]2\theta=90^\circ[/tex]
Solving:
[tex]\theta=45^\circ[/tex]
The horizontal range is maximum when θ is equal to 45°
The horizontal range is maximum for the value of angle equivalent to [tex]45^{\circ}[/tex].
Given data:
The player kicks the ball with an angle [tex]\theta[/tex] with the horizontal.
The given problem is based on the concept and fundamentals of projectile motion. In a projectile motion, the horizontal range is the horizontal distance covered by an object during its trajectory.
The mathematical expression for the horizontal range is given as,
[tex]R = \dfrac{u^{2}sin2 \theta}{g}[/tex]
Here,
R is the horizontal range.
u is the initial speed.
And g is the gravitational acceleration.
Clearly, if we need maximum horizontal range, then we need to have the maximum value of sine. So if,
[tex]\theta =45^{\circ}[/tex]
[tex]sin2 \theta=sin90[/tex] = 1
So the maximum range is,
[tex]R = \dfrac{u^{2}sin(2 \times 45^{\circ})}{g}\\\\R = \dfrac{u^{2} \times sin 90^{\circ}}{g}\\\\R = \dfrac{u^{2}}{g}[/tex]
Thus, we can conclude that the horizontal range is maximum for the value of angle equivalent to [tex]45^{\circ}[/tex].
Learn more about the projectile motion here:
https://brainly.com/question/13358421
