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Question Number 16691 by Tinkutara last updated on 25/Jun/17

$$\mathrm{The}\mathrm{horizontal}\mathrm{range}\mathrm{of}\mathrm{a}\mathrm{projectile}$$$$\mathrm{is}R\mathrm{and}\mathrm{the}\mathrm{maximum}\mathrm{height}\mathrm{attained}$$$$\mathrm{by}\mathrm{it}\mathrm{is}H.\mathrm{A}\mathrm{strong}\mathrm{wind}\mathrm{now}\mathrm{begins}\mathrm{to}$$$$\mathrm{blow}\mathrm{in}\mathrm{the}\mathrm{direction}\mathrm{of}\mathrm{horizontal}$$$$\mathrm{motion}\mathrm{of}\mathrm{projectile},\mathrm{giving}\mathrm{it}\mathrm{a}\mathrm{constant}$$$$\mathrm{horizontal}\mathrm{acceleration}\mathrm{equal}\mathrm{to}g.$$$$\mathrm{Under}\mathrm{the}\mathrm{same}\mathrm{conditions}\mathrm{of}\mathrm{projection},$$$$\mathrm{the}\mathrm{new}\mathrm{range}\mathrm{will}\mathrm{be}$$$$(g=\mathrm{acceleration}\mathrm{due}\mathrm{to}\mathrm{gravity})$$$$[\mathbf{Answer}:R+4H]$$

Answered by ajfour last updated on 25/Jun/17

$$\mathrm{time}\mathrm{of}\mathrm{flight}{\mathrm{doesn}}^{\prime }\mathrm{t}\mathrm{change},$$$$\mathrm{T}=\frac{{2\mathrm{u}}_{\mathrm{y}}}{\mathrm{g}}=\frac{2\mathrm{usin}\theta }{\mathrm{g}}$$$$\mathrm{new}\mathrm{Range}\mathrm{is}$$$${\mathrm{R}}^{\prime }=\left(\mathrm{ucos}\theta \right)\mathrm{T}+\frac{1}{2}{\mathrm{gT}}^2$$$$=\mathrm{R}+4\left[\frac{1}{2}\mathrm{g}{\left(\frac{\mathrm{T}}{2}\right)}^2\right]$$$$=\mathrm{R}+4\mathrm{H}.$$

Commented by Tinkutara last updated on 25/Jun/17

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

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