
A projectile has a range of \[50m\] and reaches a maximum height of \[10m\] calculating the angle at which the projectile is fired.
A. ${\tan ^1}\dfrac{3}{5}$
B. ${\tan ^{ - 1}}\dfrac{2}{5}$
C. ${\tan ^{ - 1}}\dfrac{3}{7}$
D. ${\tan ^{ - 1}}\dfrac{4}{5}$
Answer
497.1k+ views
Hint: The study of the motion of an object under gravity is called projectile motion. The height at which the speed of the object becomes zero from the ground is called the maximum height attained by the projectile. The formula for the range of a projectile is needed here. Then the value of the initial velocity has to be calculated. From these, the required angle can be found.
Complete step by step solution:
Let us first write the information given in the question.
Range of projectile = \[50m\], maximum height = \[10m\]
We have to calculate the angle at which the projectile is fired.
The maximum height of the projectile is given by the following formula.
$H = \dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}}$……………..(1)
Here, $u$ is the initial speed of the projectile, $g$ is the acceleration due to gravity, and $\theta $ is the angle of projection.
Now, the formula for the range of a projectile is also given below.
$R = \dfrac{{{u^2}\sin 2\theta }}{g}$
From this, we can find the value of initial velocity.
${u^2} = \dfrac{{Rg}}{{\sin 2\theta }}$
Let us substitute this value in equation (1).
$H = \left( {\dfrac{{Rg}}{{\sin 2\theta }}} \right)\dfrac{{{{\sin }^2}\theta }}{{2g}} = \dfrac{{R \times {{\sin }^2}\theta }}{{4\sin \theta cos\theta }} \Rightarrow H = \dfrac{{R\sin \theta }}{{4\cos \theta }} = \dfrac{R}{4}\tan \theta $
Let us now put the values in the above formula.
$10 = \dfrac{{50}}{4}\tan \theta \Rightarrow \tan \theta = \dfrac{4}{5}$
$\theta = {\tan ^{ - 1}}\left( {\dfrac{4}{5}} \right) \Rightarrow \theta = {57^o}$
Therefore, the angle of projection is ${57^o}$
Hence, the correct option is (D) ${\tan ^{ - 1}}\dfrac{4}{5}$.
Note:
The speed with which an object is projected is the same speed with which the object has when it reaches the ground.
The study of the projectile is done based on three quantities namely, range, time of flight, and maximum height attained by the projectile.
The range is the horizontal distance covered by the projectile from the initial point.
Time of flight is the time taken by the projectile to cover the whole path/range.
Complete step by step solution:
Let us first write the information given in the question.
Range of projectile = \[50m\], maximum height = \[10m\]
We have to calculate the angle at which the projectile is fired.
The maximum height of the projectile is given by the following formula.
$H = \dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}}$……………..(1)
Here, $u$ is the initial speed of the projectile, $g$ is the acceleration due to gravity, and $\theta $ is the angle of projection.
Now, the formula for the range of a projectile is also given below.
$R = \dfrac{{{u^2}\sin 2\theta }}{g}$
From this, we can find the value of initial velocity.
${u^2} = \dfrac{{Rg}}{{\sin 2\theta }}$
Let us substitute this value in equation (1).
$H = \left( {\dfrac{{Rg}}{{\sin 2\theta }}} \right)\dfrac{{{{\sin }^2}\theta }}{{2g}} = \dfrac{{R \times {{\sin }^2}\theta }}{{4\sin \theta cos\theta }} \Rightarrow H = \dfrac{{R\sin \theta }}{{4\cos \theta }} = \dfrac{R}{4}\tan \theta $
Let us now put the values in the above formula.
$10 = \dfrac{{50}}{4}\tan \theta \Rightarrow \tan \theta = \dfrac{4}{5}$
$\theta = {\tan ^{ - 1}}\left( {\dfrac{4}{5}} \right) \Rightarrow \theta = {57^o}$
Therefore, the angle of projection is ${57^o}$
Hence, the correct option is (D) ${\tan ^{ - 1}}\dfrac{4}{5}$.
Note:
The speed with which an object is projected is the same speed with which the object has when it reaches the ground.
The study of the projectile is done based on three quantities namely, range, time of flight, and maximum height attained by the projectile.
The range is the horizontal distance covered by the projectile from the initial point.
Time of flight is the time taken by the projectile to cover the whole path/range.
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