","comment":{"@type":"Comment","text":"Path equation is $y=x\\tan \\theta -\\dfrac{1}{2}\\dfrac{g{{x}^{2}}}{{{u}^{2}}{{\\cos }^{2}}\\theta }$ , use it to solve problem. "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$\\sqrt{3}x-10{{x}^{2}}$$","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" $$\\dfrac{1}{\\sqrt{3}}+10{{x}^{2}}$$","position":2},{"@type":"Answer","encodingFormat":"text/html","text":" $$\\dfrac{1}{\\sqrt{3}}-20{{x}^{2}}$$","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$\\sqrt{3}x-20{{x}^{2}}$$","position":1,"answerExplanation":{"@type":"Comment","text":" $$\\text{From second equation of motion, } \\\\ \\text{s=ut+}\\dfrac{1}{2}a{{t}^{2}} \\\\ \\text{here, for motion along x-axis, } \\\\ \\text{s=x, }{{\\text{u}}_{x}}\\text{=ucos}\\theta \\text{, a=0} \\\\ \\Rightarrow \\text{x=ucos}\\theta \\times \\text{t} \\\\ \\Rightarrow \\text{t=}\\dfrac{x}{u\\cos \\theta } \\\\ \\Rightarrow \\text{t=} x\\tan \\theta -\\dfrac{1}{2}\\dfrac{g{{x}^{2}}}{{{u}^{2}}{{\\cos }^{2}}\\theta } \\\\ \\text{here, for motion along y-axis, } \\\\ \\text{s=y, }{{\\text{u}}_{y}}\\text{=usin}\\theta \\text{, a=-g} \\\\ \\Rightarrow y=\\text{usin}\\theta \\times t-\\dfrac{1}{2}\\times \\left( -g\\times {{t}^{2}} \\right) \\\\ \\Rightarrow y=\\text{usin}\\theta \\times \\dfrac{x}{u\\cos \\theta }-\\dfrac{1}{2}\\times \\left( -g\\times {{\\left( \\dfrac{x}{u\\cos \\theta } \\right)}^{2}} \\right) \\\\ As\\text{ }\\theta =60{}^\\circ ,\\text{ g=-10m}{{\\text{s}}^{-2}} \\\\ \\Rightarrow y=\\sqrt{3}x-\\dfrac{1}{2}\\times \\dfrac{10\\times {{x}^{2}}}{1\\times \\dfrac{1}{4}} \\\\ \\therefore y=\\sqrt{3}x-20{{x}^{2}}$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Projectile motion Quiz 1","text":"A ball is thrown from a building of height $$H$$ with an angle $$\\theta $$ as shown in figure. Time taken by ball to reach the ground is $$ \\\\ $$ ","comment":{"@type":"Comment","text":"Take vertical displacement as $H$ and initial velocity $V$ then use the equation of motion. "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$\\sqrt{V\\sin \\theta gH}$$","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" $$\\dfrac{V\\cos \\theta \\pm \\sqrt{{{V}^{2}}{{\\cos }^{2}}\\theta +20H}}{10}$$","position":2},{"@type":"Answer","encodingFormat":"text/html","text":" $$\\dfrac{V\\sin \\theta +\\sqrt{{{V}^{2}}{{\\cos }^{2}}\\theta +20H}}{10}$$","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$\\dfrac{V\\sin \\theta \\pm \\sqrt{{{V}^{2}}{{\\sin }^{2}}\\theta +20H}}{10}$$","position":1,"answerExplanation":{"@type":"Comment","text":"
$$ \\\\ {{S}_{y}}=H,{{V}_{y}}=-V\\sin \\theta ,a=+g \\\\ \\text{(Taking downward motion }+ve\\text{)} \\\\ \\Rightarrow {{S}_{y}}={{U}_{y}}t+\\dfrac{1}{2}{{a}_{y}}{{t}^{2}} \\\\ \\Rightarrow H=-V\\sin \\theta t+5{{t}^{2}} \\\\ \\Rightarrow 5{{t}^{2}}-V\\sin \\theta -H=0 \\\\ \\Rightarrow t=\\dfrac{V\\sin \\theta \\pm \\sqrt{{{V}^{2}}{{\\sin }^{2}}\\theta +\\left( 20 \\right)H}}{10} \\\\ \\therefore t=\\dfrac{V\\sin \\theta \\pm \\sqrt{{{V}^{2}}{{\\sin }^{2}}\\theta +\\left( 20H \\right)}}{10}$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Projectile motion Quiz 1","text":" A vehicle is travelling at a speed of $$1\\,m/s$$. A man sees a child and throws chocolates for him at a speed of $$10\\,m/s$$ at a horizontal angle of $$30^\\circ $$. The distance travelled by chocolate until it reaches the child is","comment":{"@type":"Comment","text":"Take horizontal velocity of chocolate as speed of car then apply concept of projectile. "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$4.32\\,m$$","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" $$2.16\\,m$$","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" $$5\\,m$$","position":2}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$9.66\\,m$$","position":3,"answerExplanation":{"@type":"Comment","text":" $$\\text{Vertical velocity of chocolate }{{V}_{y}}=V\\sin \\theta \\\\ \\Rightarrow {{V}_{y}}=10\\times \\dfrac{1}{2} \\\\ \\Rightarrow {{V}_{y}} = 5\\text{ m/s} \\\\ \\text{Horizontal velocity of chocolate=}{{\\text{V}}_{chocolate}}+{{V}_{car}} \\\\ \\Rightarrow \\,{{V}_{x}}=V\\cos \\theta +1 \\\\ \\Rightarrow {{V}_{x}}=\\dfrac{\\sqrt{3}}{2}\\times 10+1 \\\\ \\Rightarrow {{V}_{x}}=9.66\\text{m/s} \\\\ \\text{Time of flight}=\\dfrac{2{{V}_{y}}}{g} \\\\ \\Rightarrow\\text{Time of flight}= \\dfrac{2\\times 5}{10} \\\\ \\Rightarrow\\text{Time of flight}= \\text{1 second} \\\\ \\Rightarrow \\text{Range }={{V}_{x}}\\times \\text{ }\\!\\!~\\!\\!\\text{ time of flight} \\\\ \\Rightarrow \\text{Range }=9.66\\times 1 \\\\ \\therefore \\text{Range }=9.66\\,\\text{m}$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Projectile motion Quiz 1","text":" An object of man $$m$$ is thrown from height $$H$$ with velocity of $$\\sqrt{gH}$$ $$ \\\\ $$ in case $$\\left( 1 \\right)$$ and with zero in case $$ \\\\ $$ $$\\left( 2 \\right)$$. Ratio of time taken to reach a round is $$ \\\\ $$
","comment":{"@type":"Comment","text":"Time of flight of particle depends on vertical velocity of projection. Use the formula and put the corresponding values. "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$\\sqrt{gH}:1$$","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" $$1:\\sqrt{gH}$$","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" None of these","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$1:1$$","position":2,"answerExplanation":{"@type":"Comment","text":" $$\\text{In both case }\\left( 1 \\right)\\text{ and }\\left( 2 \\right), \\\\ \\text{ the component of initial velocity along} \\\\ \\text{y-axis is zero}\\text{.} \\\\ \\text{Thus in both cases }{{u}_{f}}=0,\\text{ }{{a}_{y}}=-g,\\text{ }{{S}_{y}}=-H \\\\ \\text{Time taken}\\left( t \\right) \\\\{{S}_{y}}={{u}_{y}}t+\\dfrac{1}{2}{{a}_{y}}{{t}^{2}} \\\\ \\Rightarrow -H=0\\times t+\\dfrac{1}{2}\\times -g\\times {{t}^{2}} \\\\ \\therefore t=\\sqrt{\\dfrac{2H}{g}}$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Projectile motion Quiz 1","text":"As seen in Figure, two identical balls of mass $$m$$ are projected towards each other at the same speed $$V=10\\,m/s$$. What is the amount of time taken by them before they collide? $$ \\\\ $$
","comment":{"@type":"Comment","text":"Take relative velocity of one with respect to other and take distance to be covered equal to $10$ meter. "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$1\\sec $$","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" $$2\\sec $$","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" $$0.25\\sec $$","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$0.5\\sec $$","position":2,"answerExplanation":{"@type":"Comment","text":" $$\\text{Add in beginning} \\\\ \\text{From first equation of motion} \\\\v=u+at \\\\ \\Rightarrow a=-g\\text{ for body at ground,} \\\\While\\text{ }u=-10\\text{m/s},\\text{ a}=-g\\text{ for body at top}\\text{.} \\\\ \\Rightarrow {{V}_{ground}}=10+\\left( -10 \\right)t \\\\ \\Rightarrow {{V}_{top}}=-10+\\left( -10 \\right)t \\\\V'\\text{ is velocity of ball at ground relative to ball at top} \\\\ \\text{Relative displacement travelled}=10m \\\\ \\text{As both balls are having same acceleration,} \\\\ \\text{So no relative acceleration is present, Thus} \\\\ \\Rightarrow \\text{Displacement}=Velocity\\times Time \\\\ \\Rightarrow 10=20\\times t \\\\ \\Rightarrow t=\\dfrac{1}{2} \\\\ \\therefore t=0.5\\sec$$ ","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]}]}