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question iconCan anyone tell me where this equation came from? t = 2u / g. .... t =time, u = initial velocity, g.
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Can anyone tell me where this equation came from? t = 2u / g. t = time, u = initial velocity, g = gravity acceleration Question: I throw a cricket ball as hard as I can upwards and find that it takes 3.8s to return to the ground. Estimate the throwing speed. Answer: Use constant acceleration equation model Neglecting my height: h = u t - (1/2)gt2 Factorizing: h = 0 when t = 0 (throw) and t = 2u / g t = 3.8s, g = 9.81ms-2 So u = tg / 2 = (3.8s * 9.81ms-2) / 2 = 18.6ms-1 ------------------------------------------------- My question: Where does t = 2u / g come from?Feb 14 2021 02:46 PM

Try Solving it with these stepsThese are AI-generated tips. For the verified answer, click “Solution.pdf.”

    To understand the origin of the equation t = 2u / g, consider the following tips:

    • Review the basic equations of motion under constant acceleration.
    • Focus on the relationship between time, initial velocity, and acceleration due to gravity.
    • Analyze the scenario of an object thrown upwards and its motion characteristics.
    • Consider the symmetry of the motion: time up equals time down.
    • Derive the equation from the kinematic equations for vertical motion.
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Answered on |May 22 2021 07:34 AM

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