Physics

Physics

Problem

An object is projected vertically upwards at time t=0t = 0. Air resistance is negligible. The object passes the same point above its starting position at times 2 s and 8 s. If g=10 m s−2g = 10\text{ m s}^{-2}, what is the initial speed of the object? A. 50 B. 30 C. 25 D. 4
Intuitive explanation

The key insight is that the object's height follows a symmetric parabola—it reaches the same point on the way up and on the way down. The midpoint between these two times marks the peak, where velocity momentarily becomes zero. Use this symmetry to find when the object reaches maximum height, then apply v=u−gtv = u - gt at that turning point to work backwards to the initial speed uu.

Intuitive explanation

The key insight is that the object's height follows a symmetric parabola—it reaches the same point on the way up and on the way down. The midpoint between these two times marks the peak, where velocity momentarily becomes zero. Use this symmetry to find when the object reaches maximum height, then apply v=u−gtv = u - gt at that turning point to work backwards to the initial speed uu.

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