
In a soccer practice session the ball is kept at the center of the field 40 yards from the 10ft high goal post. A goal is attempted by kicking the ball at a speed of $64\;{\text{ft/s}}$ at an angle of ${45^\circ }$ to the horizontal. Will the ball reach the goal post?
Answer
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Hint: The time taken for the displacement of the ball can be found from the expression for the horizontal displacement. By finding the time, we can calculate the vertical displacement. The result can be compared with the height of the goal post.
Complete step by step answer:
Given the initial speed of the ball is $u = 64\;{\text{ft/s}}$, the height of goal post is $10\;{\text{ft}}$.
The angle at the ball is kicked is $\theta = {45^\circ}$. And we know that $40\;{\text{yards}} = 120\;{\text{ft}}$.
The expression for the horizontal range is given as,
$x = u\cos \theta \times t$
Where, $u$ is the initial velocity, $\theta $ is the angle and $t$ is the time taken.
Taking the horizontal range as $120\;{\text{ft}}$.
From the above expression,
$t = \dfrac{x}{{u\cos \theta }}$
Substituting the values in the above expression,
$\Rightarrow t = \dfrac{{120\;{\text{ft}}}}{{64\;{\text{ft/s}} \times \cos {{45}^\circ }}} $
$\Rightarrow t = 2.65\;{\text{s}}$
Thus the time taken for the ball for horizontal displacement $120\;{\text{ft}}$ is $2.65\;{\text{s}}$.
The expression for the vertical displacement for a time $t$ is given as,
$y = \dfrac{{\sin \theta \left( t \right) - 1}}{{2g{t^2}}}$
Substituting the values in the above expression,
$\Rightarrow y = \dfrac{{\sin 45 \times 2.65\;{\text{s}} - 1}}{{2 \times 9.8\;{\text{m/}}{{\text{s}}^{\text{2}}} \times {{\left( {2.65} \right)}^2}}}$
$\Rightarrow y = 7.08\;{\text{ft}} $
Thus the vertical displacement of the ball is $7.08\;{\text{ft}}$ at a time $2.65\;{\text{s}}$.
Thus $7.08\;{\text{ft}}$ is less than the height of the goal post, that is $10\;{\text{ft}}$. Therefore the ball reaches the goal post.
Note:
We have to note that the maximum displacement has happened when the angle is $45^\circ$. The vertical displacement is measured in sine function and the horizontal displacement is measured in cosine function.
Complete step by step answer:
Given the initial speed of the ball is $u = 64\;{\text{ft/s}}$, the height of goal post is $10\;{\text{ft}}$.
The angle at the ball is kicked is $\theta = {45^\circ}$. And we know that $40\;{\text{yards}} = 120\;{\text{ft}}$.
The expression for the horizontal range is given as,
$x = u\cos \theta \times t$
Where, $u$ is the initial velocity, $\theta $ is the angle and $t$ is the time taken.
Taking the horizontal range as $120\;{\text{ft}}$.
From the above expression,
$t = \dfrac{x}{{u\cos \theta }}$
Substituting the values in the above expression,
$\Rightarrow t = \dfrac{{120\;{\text{ft}}}}{{64\;{\text{ft/s}} \times \cos {{45}^\circ }}} $
$\Rightarrow t = 2.65\;{\text{s}}$
Thus the time taken for the ball for horizontal displacement $120\;{\text{ft}}$ is $2.65\;{\text{s}}$.
The expression for the vertical displacement for a time $t$ is given as,
$y = \dfrac{{\sin \theta \left( t \right) - 1}}{{2g{t^2}}}$
Substituting the values in the above expression,
$\Rightarrow y = \dfrac{{\sin 45 \times 2.65\;{\text{s}} - 1}}{{2 \times 9.8\;{\text{m/}}{{\text{s}}^{\text{2}}} \times {{\left( {2.65} \right)}^2}}}$
$\Rightarrow y = 7.08\;{\text{ft}} $
Thus the vertical displacement of the ball is $7.08\;{\text{ft}}$ at a time $2.65\;{\text{s}}$.
Thus $7.08\;{\text{ft}}$ is less than the height of the goal post, that is $10\;{\text{ft}}$. Therefore the ball reaches the goal post.
Note:
We have to note that the maximum displacement has happened when the angle is $45^\circ$. The vertical displacement is measured in sine function and the horizontal displacement is measured in cosine function.
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