Question

Three particles P, Q and R are thrown from the top of the tower with the same speed. P is thrown upwards, Q horizontally and R at some angle with the horizontal, they hit the ground with the speed \[{{v}_{p}},{{v}_{q}}\And {{v}_{r}}\]respectively, then the relation which holds true is?
A- \[{{v}_{p}}={{v}_{q}}={{v}_{r}}\]
B- \[{{v}_{p}}<{{v}_{q}}>{{v}_{r}}\]
C- \[{{v}_{p}}<{{v}_{q}}<{{v}_{r}}\]
D- \[{{v}_{p}}>{{v}_{q}}>{{v}_{r}}\]

Answer 1

Hint: the particles are thrown with the same initial velocity. P is thrown vertically upwards, q is thrown horizontally that is parallel to the horizontal axis and R makes some angle with the horizontal, so R is a projectile. They hit the ground with three different velocities.

Complete step by step answer:
Let the initial velocity of all the three particles be u.
Particle P is thrown up with initial velocity u, reaches maximum height and comes back to the same point with same velocity. So, \[{{v}_{p}}=u\].
\[{{v}_{y}}=\sqrt{0+2gh}\]
\[{{v}_{y}}=\sqrt{2gh}\]
Resultant is,
\[\begin{align}
  & {{v}_{Q}}=\sqrt{v_{x}^{2}+v_{y}^{2}} \\
 & {{v}_{Q}}=\sqrt{{{u}^{2}}+2gh} \\
\end{align}\]
Particle Q is thrown horizontally, so the vertical component of the initial velocity is \[u\sin 0=0\]and horizontal component is \[u\cos 0=u\].

Particle R is thrown making at some angle with the horizontal.

Using \[{{v}^{2}}-{{u}^{2}}=2gh\]
\[v=\sqrt{{{u}^{2}}+2gh}\]
For P: \[{{v}_{p}}=\sqrt{{{u}^{2}}+2gh}\]
For Q: \[{{v}_{Q}}=\sqrt{{{u}^{2}}+2gh}\]
For R: projectile with some angle come down under the influence of gravity, hence \[{{v}_{R}}=\sqrt{{{u}^{2}}+2gh}\]
Thus, \[{{v}_{p}}={{v}_{q}}={{v}_{r}}\]. So, the correct option is (A)

Note:A trajectory or flight path is the path that an object with mass in motion follows through space as a function of time. To have the maximum range the angle with the horizontal is always \[45{}^\circ \]- In projectile problems when the body is projected whether, from the ground or a certain height, there is no acceleration in the horizontal direction. There is constant acceleration in the vertical direction which is always directed downwards and this is constant. This is the acceleration due to gravity.