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This is second of our many quizzes on Projectile motion. Click on this to access more such competitive quizzes on Mechanics.

Projectile motion Quiz 2

physics

 An object is thrown with velocity $V$ from a building height of $10\,m$ below horizontal atom angle $O=30^\circ $. So that it hits at $B$, the time required to arrive at $B$ is $\dfrac{1}{2}$ times the velocity.<br><img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/3e16eaca-32c1-4e33-981f-68c616bf14ea2470342582113785791.png"/> 

 Take ${{V}_{y}}=V\sin \theta \left( -\widehat{j} \right),{{V}_{x}}=V\cos \theta \widehat{i}$, use ${{S}_{y}}=10\,m$, ${{S}_{x}}=10\,m$ and use equation of motion to solve. 

  ${{V}_{x}}=V\cos \theta =\dfrac{V\sqrt{3}}{2} \\ \Rightarrow {{S}_{x}}=10\,m \\ \Rightarrow t=\dfrac{V}{2} \\ \Rightarrow {{S}_{x}}=\left( V\cos \theta  \right)\times \left( t \right) \\ \Rightarrow {{S}_{x}}=\dfrac{{{V}^{2}}\sqrt{3}}{4} \\ \Rightarrow {{V}^{2}}=\dfrac{4\times 10}{\sqrt{3}} \\ \therefore V=2\times {{10}^{\dfrac{1}{2}}}\times {{3}^{\dfrac{-1}{4}}}$

 If a ball is thrown horizontally from a height $H$ of a tower with a certain velocity and the range covered by it is four times the height of the tower, the velocity must be

 Range $=$ horizontal velocity $\times $ time of flight, find time of flight. also for calculation take initial vertical velocity equal to zero and range and height as 4H and H respectively. 

  <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/8c4dcbad-c36d-4a57-9167-d3f74a63100e590673599752063552.png"/>$ \\ \text{Range}=4H=V\times \text{time of flight} \\{{U}_{x}}=V,{{U}_{y}}=0 \\ \Rightarrow {{S}_{y}}=H=Ut+\dfrac{1}{2}g{{t}^{2}} \\ \Rightarrow H=0+\dfrac{1}{2}g{{t}^{2}} \\ \Rightarrow t=\sqrt{\dfrac{2H}{g}} \\ \text{Range }\!\!~\!\!\text{ }=V\times t\\ \Rightarrow 4H=\sqrt{\dfrac{2H}{g}}\times V\\ \therefore V={{2}^{\dfrac{3}{2}}}{{g}^{\dfrac{1}{2}}}{{H}^{\dfrac{1}{2}}}$

 A projectile is launched from the ground at a velocity of $V$, forming an angle of $60^\circ $ with the horizontal. During the accent, the projectile makes an angle of $30^\circ $ and its velocity becomes $V^1$; the value of $V^1$ is

Velocity at any instant of projective is $\sqrt{{{V}_{x}}^{2}+{{V}_{y}}^{2}}$ and ${{V}_{x}}=U\cos \theta $, ${{V}_{y}}=U-gt$. 

 Horizontal components remain constant throughout the motion. $ \\{{V}_{x}}=V\cos \theta  \\ \Rightarrow {{V}_{x}}=\dfrac{V}{2} \\ \text{When }\,\theta =30{}^\circ  \\ \Rightarrow{{U}_{x}}={{V}^{1}}\cos 30{}^\circ  \\ \Rightarrow{{U}_{x}}={{V}_{x}} \\ \Rightarrow V\cos 60{}^\circ ={{V}^{1}}\cos 30{}^\circ \\ \Rightarrow {{V}^{1}}=\dfrac{V\cos 60{}^\circ }{\cos 30{}^\circ }\\ \therefore {{V}^{1}} =\dfrac{V}{2}\times \dfrac{2}{\sqrt{3}}$

 Two balls are fired at the same time, one at a velocity of $100\,m/s$ and the other at a velocity of $V$, In the breeze, they collide. The velocity $V$ will then be

 Calculate $V_y$ for both cases, and the vertical distance travelled by both particles is $100meter$, and their ranges overlap as they collide.

 <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/070a459a-fb55-4187-8af0-262b5250fe9b9178148108812340447.png"/>$ \\ $ Let the point of collision be $T$. So for both projectiles, the time taken to reach $T$, as well as the horizontal range $R$ till $T$ is equal. $ \\ $ As, ${{u}_{y}}=0$ for the ball at top and ${{u}_{y}}=100\sin 60{}^\circ =50\sqrt{3}m/s$  for ball at bottom. $ \\ {{V}_{y}}={{u}_{y}}+at\text{     }\left( a=-g \right) \\ \Rightarrow V{{'}_{y}}=u{{'}_{y}}+at\text{    }\left( a=-g \right) \\ $ Relative velocity of ball at ground with respect to ball at top $ =V{{'}_{y}}-{{V}_{y}}=50\sqrt{3}$ $ \\ $ Relative displacement $ = 100m$ $ \\ $ as relative acceleration is zero $( \text{both have acceleration }-g),$ $ \\ 100=50\sqrt{3}\times t \\ \Rightarrow  t=\dfrac{2}{\sqrt{3}}\sec  \\ \Rightarrow  \text{Range of }1=\text{Range of }2 \\ \Rightarrow  V\times t=100\cos 60{}^\circ \times t \\ $ Let both balls collide at point $T$. $ \\ $ Their horizontal range and time taken is the same at $T$. $ \\ $ As, $a=0$ for x-axis $ \\ $ Range of $1 = \text{Range of }2 \\ \Rightarrow  V\times t=100\cos 60{}^\circ \times t \\ \Rightarrow V=100\times \dfrac{1}{2} \\ \therefore V=50\,m/s$

A projectile is thrown at a speed of $1\text{ }m/s$ at an angle of $30{}^\circ $ to the horizontal. The projectile at the maximum height is given an extra $1\text{ }m/s$ velocity only in the horizontal direction along the motion. Which of the following parameters would differ if the projectile would not have been given the extra velocity?

The time of flight depends on the vertical component of velocity of the projectile, as well as on the maximum height. The Range of the projectile depends on the horizontal component of the velocity of the projectile.

 $\text{The initial velocity along horizontal direction is }{{u}_{x}}=u\cos 30{}^\circ =1\times \dfrac{\sqrt{3}}{2}=\dfrac{\sqrt{3}}{2}m/s\text{ }\!\!~\!\!\text{ and along }y-axis \\ {{u}_{y}}=u\sin 30{}^\circ =1\times \dfrac{1}{2}=\dfrac{1}{2}m/s \\ \text{As the projectile reaches its highest position, horizontal velocity is increased by }1m/s. \\ \text{For time of flight, horizontal vertical displacement should be zero}\text{.} \\ \Rightarrow {{S}_{y}}={{u}_{u}}t+\dfrac{1}{2}a{{t}^{2}} \\ \Rightarrow {{S}_{y}}=0,{{u}_{y}}=\dfrac{1}{2}m/s,a=-g \\ \Rightarrow 0=\dfrac{1}{2}t-\dfrac{1}{2}g{{t}^{2}} \\ \Rightarrow t=0,\dfrac{2u\sin \theta }{g} \\ \text{Thus }t=0\text{ is for initial moment, }t=\dfrac{2u\sin \theta }{g}\text{ }\!\!~\!\!\text{ is total time of flight, so it does not change}\text{.} \\\text{Maximum height also does not change as all factors remain same}\text{.} \\ \text{But as range is }{{S}_{x}}={{u}_{x}}\times \dfrac{T}{2}+{{v}_{x}}\times \dfrac{T}{2} \\ {{u}_{x}}~\text{changes at halfway point, thus the horizontal range changes} \\ \Rightarrow {{S}_{x}}=\dfrac{\sqrt{3}}{2}\times \dfrac{T}{2}+\left( \dfrac{\sqrt{3}}{2}+1 \right)\times \dfrac{T}{2} \\ \Rightarrow {{S}_{x}}=\left( \dfrac{\sqrt{3}}{2}+\dfrac{\sqrt{3}}{2}+1 \right)\times \dfrac{T}{2} \\ \Rightarrow {{S}_{x}}=\dfrac{\left( \sqrt{3}+1 \right)}{2}T \\ \Rightarrow T=\dfrac{2u\sin \theta }{g} \\ \Rightarrow T=\left( \sqrt{3}+1 \right)\dfrac{u\sin \theta }{g} \\ \Rightarrow T=\left( \sqrt{3}+1 \right)\times \dfrac{1}{2}\times \dfrac{1}{g}\theta  \\ \Rightarrow R=\dfrac{\left( \sqrt{3}+1 \right)}{2}g \\ \text{Original range would have been} \\ \Rightarrow R'=\dfrac{{{u}^{2}}\sin 2\theta }{g}=\dfrac{{{\left( 1 \right)}^{2}}\sin 60{}^\circ }{g}=\dfrac{\sqrt{3}}{2g}$ 