Question

The angle of projection at which the horizontal range and maximum height of projectile are equal is
A) \[45\]
B) \[\theta = {\tan ^{ - 1}}\left( {0.25} \right)\]
C) \[\theta = {\tan ^{ - 1}}4\;or\;\theta = 75.96\]
D) \[60\]

Answer 1

Hint: The projection angle, at which the maximum projectile height is equal to the horizontal limit, both the data must be determined. Maximum height is the maximum height obtained by an object during its projection and horizontal range is defined as the distance covered horizontally after projection on earth’s surface.

Complete step by step answer:
A projectile motion body is projected at a certain angle to a horizontal base, where the body remains in flight condition for a certain time and falls back on earth, probably known as time of flight.
Let the required projection angle is $\theta $ and the object $v$ is approximate in $m/\sec$.
We know in projectile motion
Horizontal limit range$ = \dfrac{{{u^2}\sin 2\theta }}{g}$
And maximum height$ = \dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}}$

As asked that Horizontal Range = Maximum Height
$
   \Rightarrow \dfrac{{{u^2}\sin 2\theta }}{g} = \dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}} \\
   \Rightarrow 2\sin \theta \cos \theta = \dfrac{{{{\sin }^2}\theta }}{2} \\
   \Rightarrow u\cos \theta = \sin \theta \\
   \Rightarrow \tan \theta = 4 \\
   \Rightarrow \theta = {\tan ^{ - 1}}4 \\
   \Rightarrow \theta = {75.96^o} \\
$

If the angle of projection is \[{75.96^ \circ }\], the maximum height is equal to the horizontal range.
In this condition, \[u\] represents the beginning speed extent and \[\theta \] is the angle of projection to the shot point.

Additional Information:
The body moves under the combined effect of two velocities (\[u\] and \[v\]) i.e. horizontal and vertical components of velocity. The motion in horizontal and vertical direction can be dealt separately to arrive at the equation of motion for both components of velocity.
Time of flight is the time for which body remains in air before it falls back to earth.
Time of flight $ = \dfrac{{2u\sin \theta }}{g}$

Note: In projectile motion is a type of movement where an item moves in a respectively balanced, explanatory way. The way that the item follows is called its direction. Projectile movement possibly happens when there is one power applied toward the start of the direction, after which the main impedance is from gravity. there is no increasing speed the even way. The increasing speed,
acceleration in the vertical course is only because of gravity, otherwise called free fall:
\[ax = 0,ay = - g\].