Question #1240

Sam tosses a ball horizontally off a footbridge at 3.1 m/s. How much time passes after he releases it until its speed doubles?

Expert's answer

The horizontal projection is constant while the vertical is changing according to the equation

V_{y}=V_{0y}+gt

(we chose the downward direction as positive).

V_{x} = 3.1 m/s

V_{0y} = 0;

V_{0} = 3.1 m/s

V = √(V_{x}^{2} + V_{y}^{2}) = √(3.1^{2} +100t^{2})

V^{2}/V_{0}^{2} = 4

(3.1^{2} +100t^{2})/ 3.1^{2} = 4

t =√(3.1^{2}*3 / 100) = 0.54 s

V

(we chose the downward direction as positive).

V

V

V

V = √(V

V

(3.1

t =√(3.1

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