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Question Number 16150 by Tinkutara last updated on 18/Jun/17

A projectile is fired at an angle θ with  the horizontal direction from O.  Neglecting the air friction, it hits the  ground at B after 3 seconds. What is  the height of point A from ground?  [Use g = 10 m/s^2 ]

$$\mathrm{A}\:\mathrm{projectile}\:\mathrm{is}\:\mathrm{fired}\:\mathrm{at}\:\mathrm{an}\:\mathrm{angle}\:\theta\:\mathrm{with} \\ $$$$\mathrm{the}\:\mathrm{horizontal}\:\mathrm{direction}\:\mathrm{from}\:{O}. \\ $$$$\mathrm{Neglecting}\:\mathrm{the}\:\mathrm{air}\:\mathrm{friction},\:\mathrm{it}\:\mathrm{hits}\:\mathrm{the} \\ $$$$\mathrm{ground}\:\mathrm{at}\:{B}\:\mathrm{after}\:\mathrm{3}\:\mathrm{seconds}.\:\mathrm{What}\:\mathrm{is} \\ $$$$\mathrm{the}\:\mathrm{height}\:\mathrm{of}\:\mathrm{point}\:{A}\:\mathrm{from}\:\mathrm{ground}? \\ $$$$\left[\mathrm{Use}\:{g}\:=\:\mathrm{10}\:\mathrm{m}/\mathrm{s}^{\mathrm{2}} \right] \\ $$

Commented by Tinkutara last updated on 18/Jun/17

Answered by ajfour last updated on 18/Jun/17

let time of flight = T =3s   T=((2usin θ)/g)   ⇒ usin θ = ((gT)/2) = 15m/s  OA = uT,      AB=OAsin θ   AB = (usin θ)T            = (15m/s)(3s)  =45 m .

$${let}\:{time}\:{of}\:{flight}\:=\:{T}\:=\mathrm{3}{s} \\ $$$$\:{T}=\frac{\mathrm{2}{u}\mathrm{sin}\:\theta}{{g}}\:\:\:\Rightarrow\:{u}\mathrm{sin}\:\theta\:=\:\frac{{gT}}{\mathrm{2}}\:=\:\mathrm{15}{m}/{s} \\ $$$${OA}\:=\:{uT},\:\:\:\:\:\:{AB}={OA}\mathrm{sin}\:\theta \\ $$$$\:{AB}\:=\:\left({u}\mathrm{sin}\:\theta\right){T}\: \\ $$$$\:\:\:\:\:\:\:\:\:=\:\left(\mathrm{15}{m}/{s}\right)\left(\mathrm{3}{s}\right)\:\:=\mathrm{45}\:{m}\:. \\ $$$$ \\ $$

Commented by Tinkutara last updated on 18/Jun/17

Thanks Sir!

$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$

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