
Identify which of the following is not true about projectiles launched from ground at an angle. (one or more options may be correct)
a) The horizontal velocity is constant.
b) The vertical acceleration is upward during the first half of the flight, and downward during the second half of the flight.
c) The horizontal acceleration is zero.
d) the vertical acceleration is 10m/s.
e) The time of flight can be found out by horizontal distance divided by horizontal velocity.
Answer
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Hint
When a projectile is projected, there is no external force acting on the system in a horizontal direction and so there is no acceleration and since acceleration = 0. So the velocity in horizontal direction is constant. Therefore, ${U_X} = u\cos \theta $ remains unchanged.
Complete step by step answer
a) It is indeed true that the horizontal velocity remains constant i.e. ${U_X} = u\cos \theta $ because when a projectile is projected, there is no external force acting on the system in horizontal direction and so there is no acceleration and since acceleration = 0. So the velocity in horizontal direction is constant.
b) No, this is false as vertical acceleration always acts downward due to the force of gravitation.
c) Yes, when a projectile is projected, there is no external force acting on the system in a horizontal direction and so there is no acceleration in the horizontal direction and so horizontal acceleration is zero.
d) Yes, the vertical acceleration is 10m/s which is g due to force of the weight on the object.
e) The given statement is True. The horizontal range is given by, $R = \dfrac{{2{u^2}\sin \theta \cos \theta }}{g}$ and the horizontal velocity is given by, ${U_X} = u\cos \theta $. Dividing Range by horizontal velocity, it is evident that we obtain the time of flight of the projectile as,
$\Rightarrow T = \dfrac{R}{U} = \dfrac{{\dfrac{{2{u^2}\sin \theta \cos \theta }}{g}}}{{u\cos \theta }}$
$\Rightarrow T = \dfrac{{2u\sin \theta }}{g}$
Which is the time of flight of the projectile.
Hence the answer is option (B).
Note
The range of projectile is generally given by the formula $R = \dfrac{{{u^2}sin2\theta }}{g}$ which can be simplified as $R = \dfrac{{2{u^2}\sin \theta \cos \theta }}{g}$ by using the trigonometric identity $\sin 2\theta = 2\sin \theta \cos \theta $.
When a projectile is projected, there is no external force acting on the system in a horizontal direction and so there is no acceleration and since acceleration = 0. So the velocity in horizontal direction is constant. Therefore, ${U_X} = u\cos \theta $ remains unchanged.
Complete step by step answer
a) It is indeed true that the horizontal velocity remains constant i.e. ${U_X} = u\cos \theta $ because when a projectile is projected, there is no external force acting on the system in horizontal direction and so there is no acceleration and since acceleration = 0. So the velocity in horizontal direction is constant.
b) No, this is false as vertical acceleration always acts downward due to the force of gravitation.
c) Yes, when a projectile is projected, there is no external force acting on the system in a horizontal direction and so there is no acceleration in the horizontal direction and so horizontal acceleration is zero.
d) Yes, the vertical acceleration is 10m/s which is g due to force of the weight on the object.
e) The given statement is True. The horizontal range is given by, $R = \dfrac{{2{u^2}\sin \theta \cos \theta }}{g}$ and the horizontal velocity is given by, ${U_X} = u\cos \theta $. Dividing Range by horizontal velocity, it is evident that we obtain the time of flight of the projectile as,
$\Rightarrow T = \dfrac{R}{U} = \dfrac{{\dfrac{{2{u^2}\sin \theta \cos \theta }}{g}}}{{u\cos \theta }}$
$\Rightarrow T = \dfrac{{2u\sin \theta }}{g}$
Which is the time of flight of the projectile.
Hence the answer is option (B).
Note
The range of projectile is generally given by the formula $R = \dfrac{{{u^2}sin2\theta }}{g}$ which can be simplified as $R = \dfrac{{2{u^2}\sin \theta \cos \theta }}{g}$ by using the trigonometric identity $\sin 2\theta = 2\sin \theta \cos \theta $.
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