Answer
Verified
453k+ views
Hint: In the above question it is asked to calculate the angle of projection under some given circumstances. For that we have to equate the expression for height of the projectile to twice the range(horizontal distance) covered by the projectile. Let us assume that this projection of the body takes place in vacuum i.e. no air is present that might offer air drag to the projectile.
Complete answer:
To begin with let us first draw the above projection of the projectile for better visualization.
If we see the above figure we can clearly see that the height of the projectile is twice the range of the projectile. Let us say the time taken by the projectile to reach its maximum height is T. Therefore the total time of flight is 2T. The maximum height achieved by a projectile ignoring the air resistance is given by ${{\text{H}}_{\text{M}}}\text{=}\dfrac{{{\text{U}}^{\text{2}}}\text{Si}{{\text{n}}^{\text{2}}}\text{ }\!\!\theta\!\!\text{ }}{\text{2g}}...(1)$ where U is the initial velocity, $\theta $ is the angle of projection and g is the acceleration due to gravity. The horizontal distance covered by the projectile i.e. range is given by, $\text{R=}\dfrac{{{\text{U}}^{\text{2}}}\text{Sin2 }\!\!\theta\!\!\text{ }}{\text{g}}...(2)$.We want to calculate the angle of projection for the condition i.e.
${{H}_{M}}=2R...(3)$ Hence substituting equation 1 and 2 in 3 we get,
$\dfrac{{{\text{U}}^{\text{2}}}\text{Si}{{\text{n}}^{\text{2}}}\text{ }\!\!\theta\!\!\text{ }}{\text{2g}}=\dfrac{{{\text{U}}^{\text{2}}}\text{Sin2 }\!\!\theta\!\!\text{ }}{\text{g}}$ Further after cancelling the similar terms we get,
$\begin{align}
& \dfrac{\text{Si}{{\text{n}}^{\text{2}}}\text{ }\!\!\theta\!\!\text{ }}{\text{2}}=\text{Sin2 }\!\!\theta\!\!\text{ , since Sin2 }\!\!\theta\!\!\text{ =2Sin }\!\!\theta\!\!\text{ Cos }\!\!\theta\!\!\text{ } \\
& \dfrac{\text{Si}{{\text{n}}^{\text{2}}}\text{ }\!\!\theta\!\!\text{ }}{\text{2}}=\text{2Sin }\!\!\theta\!\!\text{ Cos }\!\!\theta\!\!\text{ } \\
& \dfrac{\text{Sin }\!\!\theta\!\!\text{ }}{\text{Cos }\!\!\theta\!\!\text{ }}=4 \\
\end{align}$
Since $\dfrac{\text{Sin }\!\!\theta\!\!\text{ }}{\text{Cos }\!\!\theta\!\!\text{ }}=\tan \theta $
$\begin{align}
& \text{Tan }\!\!\theta\!\!\text{ =4 hence} \\
& \text{ }\!\!\theta\!\!\text{ =Ta}{{\text{n}}^{\text{-1}}}\text{4=36}\text{.8}{{\text{6}}^{\text{o}}} \\
\end{align}$
Hence the angle of projection should be 36.86 degrees.
Note:
In the real world the angle of projection required to satisfy the above condition should have been slightly greater as the range would have been less due to air resistance. One more important fact is that for a given range the maximum height of the projectile can have two different values. This depends on the angle of projection.
Complete answer:
To begin with let us first draw the above projection of the projectile for better visualization.
If we see the above figure we can clearly see that the height of the projectile is twice the range of the projectile. Let us say the time taken by the projectile to reach its maximum height is T. Therefore the total time of flight is 2T. The maximum height achieved by a projectile ignoring the air resistance is given by ${{\text{H}}_{\text{M}}}\text{=}\dfrac{{{\text{U}}^{\text{2}}}\text{Si}{{\text{n}}^{\text{2}}}\text{ }\!\!\theta\!\!\text{ }}{\text{2g}}...(1)$ where U is the initial velocity, $\theta $ is the angle of projection and g is the acceleration due to gravity. The horizontal distance covered by the projectile i.e. range is given by, $\text{R=}\dfrac{{{\text{U}}^{\text{2}}}\text{Sin2 }\!\!\theta\!\!\text{ }}{\text{g}}...(2)$.We want to calculate the angle of projection for the condition i.e.
${{H}_{M}}=2R...(3)$ Hence substituting equation 1 and 2 in 3 we get,
$\dfrac{{{\text{U}}^{\text{2}}}\text{Si}{{\text{n}}^{\text{2}}}\text{ }\!\!\theta\!\!\text{ }}{\text{2g}}=\dfrac{{{\text{U}}^{\text{2}}}\text{Sin2 }\!\!\theta\!\!\text{ }}{\text{g}}$ Further after cancelling the similar terms we get,
$\begin{align}
& \dfrac{\text{Si}{{\text{n}}^{\text{2}}}\text{ }\!\!\theta\!\!\text{ }}{\text{2}}=\text{Sin2 }\!\!\theta\!\!\text{ , since Sin2 }\!\!\theta\!\!\text{ =2Sin }\!\!\theta\!\!\text{ Cos }\!\!\theta\!\!\text{ } \\
& \dfrac{\text{Si}{{\text{n}}^{\text{2}}}\text{ }\!\!\theta\!\!\text{ }}{\text{2}}=\text{2Sin }\!\!\theta\!\!\text{ Cos }\!\!\theta\!\!\text{ } \\
& \dfrac{\text{Sin }\!\!\theta\!\!\text{ }}{\text{Cos }\!\!\theta\!\!\text{ }}=4 \\
\end{align}$
Since $\dfrac{\text{Sin }\!\!\theta\!\!\text{ }}{\text{Cos }\!\!\theta\!\!\text{ }}=\tan \theta $
$\begin{align}
& \text{Tan }\!\!\theta\!\!\text{ =4 hence} \\
& \text{ }\!\!\theta\!\!\text{ =Ta}{{\text{n}}^{\text{-1}}}\text{4=36}\text{.8}{{\text{6}}^{\text{o}}} \\
\end{align}$
Hence the angle of projection should be 36.86 degrees.
Note:
In the real world the angle of projection required to satisfy the above condition should have been slightly greater as the range would have been less due to air resistance. One more important fact is that for a given range the maximum height of the projectile can have two different values. This depends on the angle of projection.
Recently Updated Pages
what is the correct chronological order of the following class 10 social science CBSE
Which of the following was not the actual cause for class 10 social science CBSE
Which of the following statements is not correct A class 10 social science CBSE
Which of the following leaders was not present in the class 10 social science CBSE
Garampani Sanctuary is located at A Diphu Assam B Gangtok class 10 social science CBSE
Which one of the following places is not covered by class 10 social science CBSE
Trending doubts
Which are the Top 10 Largest Countries of the World?
How do you graph the function fx 4x class 9 maths CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
The only snake that builds a nest is a Krait b King class 11 biology CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
In Indian rupees 1 trillion is equal to how many c class 8 maths CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
Why is there a time difference of about 5 hours between class 10 social science CBSE
Which places in India experience sunrise first and class 9 social science CBSE