Physics
A ball is thrown directly downward, with an initial speed of 6.10 m/s, from a height of 31.0 m. After what interval does the ball strike the ground?
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The interval for the ball to strike the ground can be calculated using the formula: t = -(v + √v^2 + 2ad)/a. In this equation, v is the initial velocity (6.10 m/s), a is the acceleration of gravity (9.8 m/s^2), and d is the initial displacement (31.0 m). Plugging these values into the equation yields: t = -(6.10 + √(6.10^2 + 2*(9.8)*(31.0))) / (9.8) t = -(6.10 + √380.76) / (9.8) t = -2.947 seconds Therefore, the ball will strike the ground after an interval of 2.947 seconds.
