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12 m = 10 m v 0 sin 40° t (– 980 m/s2)t2 We can use the first equation to eliminate v 0t from the second and solve for t, which gives t = 367 s When this value is used in the first equation, we get v 0 = 33 m/sExample 7 A rock is released from a height of 196mDetermine (a) the time it spends in air and (b) its speed just before striking the ground Solution Since the rock is not thrown and is only released from rest, the initial speed, v i = 0 If the ( y) axis is taken to be downward, then g = 98m/s 2, and the equation of motion,🔥 K T M L O V E R B O Y 🔥 (@hamid_k_t_m_boy) has created a short video on TikTok with music original sound ️K T M 🔥 L O V E R 🔥 B O Y #bikel #ktmlover #baikuntukharimu #bikelover #backbenchersonflipkart #ktmrc390

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About Press Copyright Contact us Creators Advertise Developers Terms Privacy Policy & Safety How works Test new features Press Copyright Contact us CreatorsFor projectile motion we have x = v cosθ t, y = v sinθ t ½gt 2 Details of the calculation (a) The distance between the rock and the launch point is D = (x 2 y 2) ½ = (v 2 t 2 g 2 t 4 /4 vgt 3 sinθ) ½ For 0 < θ < 90 o and v > 0 the distance D increases initially To find extrema for D we set dD 2 /dt = t(g 2 t 2 3vgt sinθS= ut 1 2 at 2;




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