Question #116330

A 0.5 kg ball is thrown upwards with an initial velocity of 6 m/s. How high will the ball reach before it starts falling again?

Expert's answer

Equations of motion of the ball are "v(t) = v_0 - g t" and "y(t) = v_0 t - \\frac{g t^2}{2}".

When the ball reaches its highest point, the speed is zero, hence "v(t_s) = 0 \\Rightarrow t_s = \\frac{v_0}{g}".

"t_s" is the time it took to reach the highest point. Therefore, "y" coordinate at that moment of time is how high will the ball reach before it starts falling again: "H = y(t_s) = \\frac{v_0^2}{g} - \\frac{v_0^2}{2 g} = \\frac{v_0^2}{2 g} \\approx 1.83 m".

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