A body is thrown vertically upward with speed 40 m/s. Distance travelled in last second of ascent (g=10)

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Team JKExamLibrary · Accepted Answer

Time of ascent = u/g = 4 s. Distance in last second (4th) = ½g(1)²? Actually, during last second of upward journey, velocity becomes zero. Distance = ½g(1)² = 5 m. Formula: s = u - g/2 (2n-1) with n=4? Let's compute: s_nth = u - (g/2)(2n-1). For n=4 (4th second), s = 40 - 5×7 = 40 - 35 = 5 m. Yes.

A body is thrown vertically upward with speed 40 m/s. Distance travelled in last second of ascent (g=10) MCQ with Answer and Explanation

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