Question Number 15591 by Tinkutara last updated on 12/Jun/17
$$\mathrm{A}\:\mathrm{projectile}\:\mathrm{projected}\:\mathrm{from}\:\mathrm{the}\:\mathrm{ground} \\ $$$$\mathrm{has}\:\mathrm{its}\:\mathrm{direction}\:\mathrm{of}\:\mathrm{motion}\:\mathrm{making}\:\mathrm{an} \\ $$$$\mathrm{angle}\:\frac{\pi}{\mathrm{4}}\:\mathrm{with}\:\mathrm{the}\:\mathrm{horizontal}\:\mathrm{at}\:\mathrm{a}\:\mathrm{height} \\ $$$$\mathrm{40}\:\mathrm{m}.\:\mathrm{Its}\:\mathrm{initial}\:\mathrm{velocity}\:\mathrm{of}\:\mathrm{projection}\:\mathrm{is} \\ $$$$\mathrm{50}\:\mathrm{m}/\mathrm{s},\:\mathrm{the}\:\mathrm{angle}\:\mathrm{of}\:\mathrm{projection}\:\mathrm{is}? \\ $$
Answered by mrW1 last updated on 12/Jun/17
![u=50m/s u_x =50cos α u_y =50sin α y=u_y t−(1/2)gt^2 x=u_x t tan θ=(dy/dx)=(dy/dt)×(1/(dx/dt))=((u_y −gt)/u_x ) at y=40m: ϑ=(π/4) tan (π/4)=((u_y −gt)/u_x )=1 t=((u_y −u_x )/g) y=u_y ((u_y −u_x )/g)−(g/2)(((u_y −u_x )/g))^2 =40 ((u_y −u_x )/g)[u_y −((u_y −u_x )/2)]=40 ((u_y −u_x )/g)[((u_y +u_x )/2)]=40 u_y ^2 −u_x ^2 =2×40g u^2 (sin^2 α−cos^2 α)=80g −u^2 cos 2α=80g cos 2α=−((80g)/u^2 )=−((800)/(50^2 ))=−0.32 ⇒2α=108.7° ⇒α=54.3°](https://www.tinkutara.com/question/Q15596.png)
$$\mathrm{u}=\mathrm{50m}/\mathrm{s} \\ $$$$\mathrm{u}_{\mathrm{x}} =\mathrm{50cos}\:\alpha \\ $$$$\mathrm{u}_{\mathrm{y}} =\mathrm{50sin}\:\alpha \\ $$$$\mathrm{y}=\mathrm{u}_{\mathrm{y}} \mathrm{t}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{gt}^{\mathrm{2}} \\ $$$$\mathrm{x}=\mathrm{u}_{\mathrm{x}} \mathrm{t} \\ $$$$\mathrm{tan}\:\theta=\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dy}}{\mathrm{dt}}×\frac{\mathrm{1}}{\frac{\mathrm{dx}}{\mathrm{dt}}}=\frac{\mathrm{u}_{\mathrm{y}} −\mathrm{gt}}{\mathrm{u}_{\mathrm{x}} } \\ $$$$\mathrm{at}\:\mathrm{y}=\mathrm{40m}:\:\vartheta=\frac{\pi}{\mathrm{4}} \\ $$$$\mathrm{tan}\:\frac{\pi}{\mathrm{4}}=\frac{\mathrm{u}_{\mathrm{y}} −\mathrm{gt}}{\mathrm{u}_{\mathrm{x}} }=\mathrm{1} \\ $$$$\mathrm{t}=\frac{\mathrm{u}_{\mathrm{y}} −\mathrm{u}_{\mathrm{x}} }{\mathrm{g}} \\ $$$$\mathrm{y}=\mathrm{u}_{\mathrm{y}} \frac{\mathrm{u}_{\mathrm{y}} −\mathrm{u}_{\mathrm{x}} }{\mathrm{g}}−\frac{\mathrm{g}}{\mathrm{2}}\left(\frac{\mathrm{u}_{\mathrm{y}} −\mathrm{u}_{\mathrm{x}} }{\mathrm{g}}\right)^{\mathrm{2}} =\mathrm{40} \\ $$$$\frac{\mathrm{u}_{\mathrm{y}} −\mathrm{u}_{\mathrm{x}} }{\mathrm{g}}\left[\mathrm{u}_{\mathrm{y}} −\frac{\mathrm{u}_{\mathrm{y}} −\mathrm{u}_{\mathrm{x}} }{\mathrm{2}}\right]=\mathrm{40} \\ $$$$\frac{\mathrm{u}_{\mathrm{y}} −\mathrm{u}_{\mathrm{x}} }{\mathrm{g}}\left[\frac{\mathrm{u}_{\mathrm{y}} +\mathrm{u}_{\mathrm{x}} }{\mathrm{2}}\right]=\mathrm{40} \\ $$$$\mathrm{u}_{\mathrm{y}} ^{\mathrm{2}} −\mathrm{u}_{\mathrm{x}} ^{\mathrm{2}} =\mathrm{2}×\mathrm{40g} \\ $$$$\mathrm{u}^{\mathrm{2}} \left(\mathrm{sin}^{\mathrm{2}} \:\alpha−\mathrm{cos}^{\mathrm{2}} \:\alpha\right)=\mathrm{80g} \\ $$$$−\mathrm{u}^{\mathrm{2}} \mathrm{cos}\:\mathrm{2}\alpha=\mathrm{80g} \\ $$$$\mathrm{cos}\:\mathrm{2}\alpha=−\frac{\mathrm{80g}}{\mathrm{u}^{\mathrm{2}} }=−\frac{\mathrm{800}}{\mathrm{50}^{\mathrm{2}} }=−\mathrm{0}.\mathrm{32} \\ $$$$\Rightarrow\mathrm{2}\alpha=\mathrm{108}.\mathrm{7}° \\ $$$$\Rightarrow\alpha=\mathrm{54}.\mathrm{3}° \\ $$
Commented by Tinkutara last updated on 12/Jun/17
$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$