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Question Number 15405 by Tinkutara last updated on 10/Jun/17
A body is projected at time t = 0 from a certain point on a planet surface with a certain velocity at a certain angle with the planet′s surface (assumed horizontal). The horizontal and vertical displacement x and y in metre are related to time as x = 10(√3)t and y = 10t − 4t^2 . Find vertical component of velocity of the particle when it is at a height half of the maximum height attained.
$$\mathrm{A}\:\mathrm{body}\:\mathrm{is}\:\mathrm{projected}\:\mathrm{at}\:\mathrm{time}\:{t}\:=\:\mathrm{0}\:\mathrm{from}\:\mathrm{a} \\ $$$$\mathrm{certain}\:\mathrm{point}\:\mathrm{on}\:\mathrm{a}\:\mathrm{planet}\:\mathrm{surface}\:\mathrm{with} \\ $$$$\mathrm{a}\:\mathrm{certain}\:\mathrm{velocity}\:\mathrm{at}\:\mathrm{a}\:\mathrm{certain}\:\mathrm{angle} \\ $$$$\mathrm{with}\:\mathrm{the}\:\mathrm{planet}'\mathrm{s}\:\mathrm{surface}\:\left(\mathrm{assumed}\right. \\ $$$$\left.\mathrm{horizontal}\right).\:\mathrm{The}\:\mathrm{horizontal}\:\mathrm{and}\:\mathrm{vertical} \\ $$$$\mathrm{displacement}\:{x}\:\mathrm{and}\:{y}\:\mathrm{in}\:\mathrm{metre}\:\mathrm{are} \\ $$$$\mathrm{related}\:\mathrm{to}\:\mathrm{time}\:\mathrm{as}\:{x}\:=\:\mathrm{10}\sqrt{\mathrm{3}}{t}\:\mathrm{and} \\ $$$${y}\:=\:\mathrm{10}{t}\:−\:\mathrm{4}{t}^{\mathrm{2}} .\:\mathrm{Find}\:\mathrm{vertical}\:\mathrm{component} \\ $$$$\mathrm{of}\:\mathrm{velocity}\:\mathrm{of}\:\mathrm{the}\:\mathrm{particle}\:\mathrm{when}\:\mathrm{it}\:\mathrm{is}\:\mathrm{at}\:\mathrm{a} \\ $$$$\mathrm{height}\:\mathrm{half}\:\mathrm{of}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{height} \\ $$$$\mathrm{attained}. \\ $$
Answered by ajfour last updated on 10/Jun/17
As y=u_y t+(1/2)a_y t^2 =10t−4t^2 ⇒ u_y =10m/s, a_y =−8m/s^2 v_y =u_y −at=10−8t ⇒ v_y =0 at t_(1/2) =(5/4)s height is maximum then, H=u_y t_(1/2) −(1/2)a_y t_(1/2) ^2 =10×(5/4)−(1/2)×8×((25)/(16))= ((25)/4)m. when y=(H/2) , let y component of velocity then be v_1 : v_1 ^2 −u_y ^2 =2a_y ((H/2)) v_1 ^2 =100−2×8×((25)/8)=50m^2 /s^2 v_1 =±5(√2)m/s .
$${As}\:\:\:\:\:\:{y}={u}_{{y}} {t}+\frac{\mathrm{1}}{\mathrm{2}}{a}_{{y}} {t}^{\mathrm{2}} =\mathrm{10}{t}−\mathrm{4}{t}^{\mathrm{2}} \\ $$$$\Rightarrow\:{u}_{{y}} =\mathrm{10}{m}/{s},\:\:\:{a}_{{y}} =−\mathrm{8}{m}/{s}^{\mathrm{2}} \\ $$$${v}_{{y}} ={u}_{{y}} −{at}=\mathrm{10}−\mathrm{8}{t} \\ $$$$\Rightarrow\:{v}_{{y}} =\mathrm{0}\:{at}\:{t}_{\mathrm{1}/\mathrm{2}} =\frac{\mathrm{5}}{\mathrm{4}}{s} \\ $$$${height}\:{is}\:{maximum}\:{then}, \\ $$$${H}={u}_{{y}} {t}_{\mathrm{1}/\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}{a}_{{y}} {t}_{\mathrm{1}/\mathrm{2}} ^{\mathrm{2}} \\ $$$$\:\:\:\:=\mathrm{10}×\frac{\mathrm{5}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{8}×\frac{\mathrm{25}}{\mathrm{16}}=\:\frac{\mathrm{25}}{\mathrm{4}}{m}. \\ $$$${when}\:{y}=\frac{{H}}{\mathrm{2}}\:\:,\:{let}\:{y}\:{component} \\ $$$${of}\:{velocity}\:{then}\:{be}\:\boldsymbol{{v}}_{\mathrm{1}} \:\:: \\ $$$${v}_{\mathrm{1}} ^{\mathrm{2}} −{u}_{{y}} ^{\mathrm{2}} =\mathrm{2}{a}_{{y}} \left(\frac{{H}}{\mathrm{2}}\right) \\ $$$${v}_{\mathrm{1}} ^{\mathrm{2}} =\mathrm{100}−\mathrm{2}×\mathrm{8}×\frac{\mathrm{25}}{\mathrm{8}}=\mathrm{50}{m}^{\mathrm{2}} /{s}^{\mathrm{2}} \\ $$$$\:\:\boldsymbol{{v}}_{\mathrm{1}} =\pm\mathrm{5}\sqrt{\mathrm{2}}\boldsymbol{{m}}/\boldsymbol{{s}}\:. \\ $$
Commented by Tinkutara last updated on 10/Jun/17
Will it be only 5(√2) or ± 5(√2) m/s?
$$\mathrm{Will}\:\mathrm{it}\:\mathrm{be}\:\mathrm{only}\:\mathrm{5}\sqrt{\mathrm{2}}\:\mathrm{or}\:\pm\:\mathrm{5}\sqrt{\mathrm{2}}\:\mathrm{m}/\mathrm{s}? \\ $$
Commented by ajfour last updated on 10/Jun/17
can you find distance travelled by projectile. u=5m/s, u_x =3m/s, g=10m/s^2 . distance travelled=?
$${can}\:{you}\:{find}\:{distance}\:{travelled} \\ $$$${by}\:{projectile}.\:{u}=\mathrm{5}{m}/{s},\:{u}_{{x}} =\mathrm{3}{m}/{s}, \\ $$$${g}=\mathrm{10}{m}/{s}^{\mathrm{2}} .\:{distance}\:{travelled}=? \\ $$
Commented by Tinkutara last updated on 10/Jun/17
Please check my answer comes out to be approximately 3 m.
$$\mathrm{Please}\:\mathrm{check}\:\mathrm{my}\:\mathrm{answer}\:\mathrm{comes}\:\mathrm{out}\:\mathrm{to} \\ $$$$\mathrm{be}\:\mathrm{approximately}\:\mathrm{3}\:\mathrm{m}. \\ $$
Commented by ajfour last updated on 10/Jun/17
i can check your method even if you describe in short..
$${i}\:{can}\:{check}\:{your}\:{method}\:{even}\:{if} \\ $$$${you}\:{describe}\:{in}\:{short}.. \\ $$
Commented by Tinkutara last updated on 10/Jun/17
Is the distance travelled by the particle the same as length of the parabola in the 1^(st) quadrant formed by the equation of its trajectory?
$$\mathrm{Is}\:\mathrm{the}\:\mathrm{distance}\:\mathrm{travelled}\:\mathrm{by}\:\mathrm{the}\:\mathrm{particle} \\ $$$$\mathrm{the}\:\mathrm{same}\:\mathrm{as}\:\mathrm{length}\:\mathrm{of}\:\mathrm{the}\:\mathrm{parabola}\:\mathrm{in} \\ $$$$\mathrm{the}\:\mathrm{1}^{\mathrm{st}} \:\mathrm{quadrant}\:\mathrm{formed}\:\mathrm{by}\:\mathrm{the} \\ $$$$\mathrm{equation}\:\mathrm{of}\:\mathrm{its}\:\mathrm{trajectory}? \\ $$
Commented by ajfour last updated on 10/Jun/17
yes , true .
$${yes}\:,\:{true}\:. \\ $$
Commented by Tinkutara last updated on 10/Jun/17
Then we will calculate the length of the curve using (ds)^2 = (dx)^2 + (dy)^2 . ds = (√((dx)^2 + (dy)^2 )) Here u cos θ = 3 and u sin θ = 4. Time of flight = ((2u sin θ)/g) = (4/5) Displacement along x = 3t and y = 4t − 5t^2 dx = 3dt and dy = (4 − 10t)dt ds = (√(3^2 + (4 − 10t)^2 )) dt ∫_0 ^s ds = ∫_0 ^(4/5) (√(100t^2 − 80t + 25)) dt , which will give s = 2 + (9/(10)) ln 3 ≈ 2.99 m
$$\mathrm{Then}\:\mathrm{we}\:\mathrm{will}\:\mathrm{calculate}\:\mathrm{the}\:\mathrm{length}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{curve}\:\mathrm{using}\:\left({ds}\right)^{\mathrm{2}} \:=\:\left({dx}\right)^{\mathrm{2}} \:+\:\left({dy}\right)^{\mathrm{2}} . \\ $$$${ds}\:=\:\sqrt{\left({dx}\right)^{\mathrm{2}} \:+\:\left({dy}\right)^{\mathrm{2}} } \\ $$$$\mathrm{Here}\:{u}\:\mathrm{cos}\:\theta\:=\:\mathrm{3}\:\mathrm{and}\:{u}\:\mathrm{sin}\:\theta\:=\:\mathrm{4}. \\ $$$$\mathrm{Time}\:\mathrm{of}\:\mathrm{flight}\:=\:\frac{\mathrm{2}{u}\:\mathrm{sin}\:\theta}{{g}}\:=\:\frac{\mathrm{4}}{\mathrm{5}} \\ $$$$\mathrm{Displacement}\:\mathrm{along}\:{x}\:=\:\mathrm{3}{t}\:\mathrm{and} \\ $$$${y}\:=\:\mathrm{4}{t}\:−\:\mathrm{5}{t}^{\mathrm{2}} \\ $$$${dx}\:=\:\mathrm{3}{dt}\:\mathrm{and}\:{dy}\:=\:\left(\mathrm{4}\:−\:\mathrm{10}{t}\right){dt} \\ $$$${ds}\:=\:\sqrt{\mathrm{3}^{\mathrm{2}} \:+\:\left(\mathrm{4}\:−\:\mathrm{10}{t}\right)^{\mathrm{2}} }\:{dt} \\ $$$$\int_{\mathrm{0}} ^{{s}} {ds}\:=\:\int_{\mathrm{0}} ^{\frac{\mathrm{4}}{\mathrm{5}}} \sqrt{\mathrm{100}{t}^{\mathrm{2}} \:−\:\mathrm{80}{t}\:+\:\mathrm{25}}\:{dt}\:, \\ $$$$\mathrm{which}\:\mathrm{will}\:\mathrm{give}\:{s}\:=\:\mathrm{2}\:+\:\frac{\mathrm{9}}{\mathrm{10}}\:\mathrm{ln}\:\mathrm{3}\:\approx\:\mathrm{2}.\mathrm{99}\:\mathrm{m} \\ $$
Commented by ajfour last updated on 10/Jun/17
All the best for JEE entrance !
$${All}\:{the}\:{best}\:{for}\:{JEE}\:{entrance}\:! \\ $$
Commented by Tinkutara last updated on 11/Jun/17
But I am just in Class 11 now.
$$\mathrm{But}\:\mathrm{I}\:\mathrm{am}\:\mathrm{just}\:\mathrm{in}\:\mathrm{Class}\:\mathrm{11}\:\mathrm{now}. \\ $$
Commented by prakash jain last updated on 12/Jun/17
So u will appear in JEE2019? Are u attending any coaching?
$$\mathrm{So}\:\mathrm{u}\:\mathrm{will}\:\mathrm{appear}\:\mathrm{in}\:\mathrm{JEE2019}? \\ $$$$\mathrm{Are}\:\mathrm{u}\:\mathrm{attending}\:\mathrm{any}\:\mathrm{coaching}? \\ $$
Commented by Tinkutara last updated on 12/Jun/17
Yes in 2019. I am attending my regular coaching classes.
$$\mathrm{Yes}\:\mathrm{in}\:\mathrm{2019}.\:\mathrm{I}\:\mathrm{am}\:\mathrm{attending}\:\mathrm{my}\:\mathrm{regular} \\ $$$$\mathrm{coaching}\:\mathrm{classes}. \\ $$
Commented by prakash jain last updated on 12/Jun/17
I also did B. Tech. from IIT Chennai in 1993. Now my son is also preparing for JEE 2019.
$$\mathrm{I}\:\mathrm{also}\:\mathrm{did}\:\mathrm{B}.\:\mathrm{Tech}.\:\mathrm{from}\:\mathrm{IIT}\:\mathrm{Chennai} \\ $$$$\mathrm{in}\:\mathrm{1993}.\: \\ $$$$\mathrm{Now}\:\mathrm{my}\:\mathrm{son}\:\mathrm{is}\:\mathrm{also}\:\mathrm{preparing}\:\mathrm{for} \\ $$$$\mathrm{JEE}\:\mathrm{2019}. \\ $$
Commented by Tinkutara last updated on 12/Jun/17
Had he joined any coaching?
$$\mathrm{Had}\:\mathrm{he}\:\mathrm{joined}\:\mathrm{any}\:\mathrm{coaching}? \\ $$
Commented by prakash jain last updated on 12/Jun/17
Yes. FIITJEE.
$$\mathrm{Yes}.\:\mathrm{FIITJEE}. \\ $$
Commented by Tinkutara last updated on 12/Jun/17
I had joined Aakash. What was his Class 10^(th) board result? He is in which centre of FIITJEE?
$$\mathrm{I}\:\mathrm{had}\:\mathrm{joined}\:\mathrm{Aakash}.\:\mathrm{What}\:\mathrm{was}\:\mathrm{his} \\ $$$$\mathrm{Class}\:\mathrm{10}^{\mathrm{th}} \:\mathrm{board}\:\mathrm{result}?\:\mathrm{He}\:\mathrm{is}\:\mathrm{in}\:\mathrm{which} \\ $$$$\mathrm{centre}\:\mathrm{of}\:\mathrm{FIITJEE}? \\ $$
Commented by mrW1 last updated on 13/Jun/17
You all talk about JEE. What′s that? I never had it and even heard of it.
$$\mathrm{You}\:\mathrm{all}\:\mathrm{talk}\:\mathrm{about}\:\mathrm{JEE}.\:\mathrm{What}'\mathrm{s}\:\mathrm{that}? \\ $$$$\mathrm{I}\:\mathrm{never}\:\mathrm{had}\:\mathrm{it}\:\mathrm{and}\:\mathrm{even}\:\mathrm{heard}\:\mathrm{of}\:\mathrm{it}. \\ $$
Commented by Tinkutara last updated on 13/Jun/17
Sir in India JEE is an entrance exam from which we can get to engineering colleges. It is meant for Class 12 students.
$$\mathrm{Sir}\:\mathrm{in}\:\mathrm{India}\:\mathrm{JEE}\:\mathrm{is}\:\mathrm{an}\:\mathrm{entrance}\:\mathrm{exam} \\ $$$$\mathrm{from}\:\mathrm{which}\:\mathrm{we}\:\mathrm{can}\:\mathrm{get}\:\mathrm{to}\:\mathrm{engineering} \\ $$$$\mathrm{colleges}.\:\mathrm{It}\:\mathrm{is}\:\mathrm{meant}\:\mathrm{for}\:\mathrm{Class}\:\mathrm{12} \\ $$$$\mathrm{students}. \\ $$
Commented by mrW1 last updated on 13/Jun/17
thanks! then i wish you all success in jee!
$$\mathrm{thanks}!\:\mathrm{then}\:\mathrm{i}\:\mathrm{wish}\:\mathrm{you}\:\mathrm{all}\:\mathrm{success}\:\mathrm{in} \\ $$$$\mathrm{jee}! \\ $$

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