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Question: A projectile has a range of \[50m\] and reaches a maximum height of \[10m\] calculating the angle at...

A projectile has a range of 50m50m and reaches a maximum height of 10m10m calculating the angle at which the projectile is fired.
A. tan135{\tan ^1}\dfrac{3}{5}
B. tan125{\tan ^{ - 1}}\dfrac{2}{5}
C. tan137{\tan ^{ - 1}}\dfrac{3}{7}
D. tan145{\tan ^{ - 1}}\dfrac{4}{5}

Explanation

Solution

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 = 50m50m, maximum height = 10m10m
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=u2sin2θ2gH = \dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}}……………..(1)
Here, uu is the initial speed of the projectile, gg is the acceleration due to gravity, and θ\theta is the angle of projection.
Now, the formula for the range of a projectile is also given below.
R=u2sin2θgR = \dfrac{{{u^2}\sin 2\theta }}{g}
From this, we can find the value of initial velocity.
u2=Rgsin2θ{u^2} = \dfrac{{Rg}}{{\sin 2\theta }}
Let us substitute this value in equation (1).
H=(Rgsin2θ)sin2θ2g=R×sin2θ4sinθcosθH=Rsinθ4cosθ=R4tanθH = \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=504tanθtanθ=4510 = \dfrac{{50}}{4}\tan \theta \Rightarrow \tan \theta = \dfrac{4}{5}
θ=tan1(45)θ=57o\theta = {\tan ^{ - 1}}\left( {\dfrac{4}{5}} \right) \Rightarrow \theta = {57^o}
Therefore, the angle of projection is 57o{57^o}
Hence, the correct option is (D) tan145{\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.