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Q.Find the expression for the time of flight, maximum height and horizontal range of a projectile.

Bihar BsebBihar Board Intermediate 1st Year 2024Subjective· 5mImportance★★★★★
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T=2usin⁡θgT=\dfrac{2u\sin\theta}{g}, H=u2sin⁡2θ2gH=\dfrac{u^2\sin^2\theta}{2g}, R=u2sin⁡2θgR=\dfrac{u^2\sin2\theta}{g}.

A projectile is launched with initial speed uu at angle θ\theta above the horizontal. Taking the launch point as origin, with xx horizontal and yy vertical (positive upward, gravity gg acting downward):

Initial velocity components: ux=ucos⁡θu_x=u\cos\theta (constant throughout, no horizontal force), uy=usin⁡θu_y=u\sin\theta.

Time of flight TT: The vertical motion is y=uyt−12gt2y=u_y t-\tfrac12gt^2. The projectile lands when y=0y=0 again (same launch height): uyT−12gT2=0⇒T(uy−12gT)=0u_y T-\tfrac12gT^2=0 \Rightarrow T\left(u_y-\tfrac12gT\right)=0. The non-trivial solution is T=2uyg=2usin⁡θgT=\dfrac{2u_y}{g}=\dfrac{2u\sin\theta}{g}.

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