A ball is thrown upwards from a bridge at 1.9 m s-1. How much time later does it hit the water 6.7 m below? (Give your answer to one decimal place.)

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Begin by identifying the motion scenario: a ball is projected upward from a height with initial vertical velocity u = +1.9 m/s (positive upward), and it falls a vertical distance of 6.7 m to the water below.

Using the vertical motion equation with upward as positive and acceleration a = -g = -9.8 m/s^2: y(t) = u t + (1/2) a t^2, where y is the vertical displacement from the starting point.

We want the displacement to reach -6.7 m (since the water is 6.7 m below the launch point): -6.7 = (1.9) t + (1/2)(-9.8) t^2.

This simplifies to: -6.7 = 1.9 t - 4.9 t^2, which can be rearranged to standard quadratic form: 4.9 t^2 - 1.9 t - 6.7 = 0.

Now solve the quadratic 4.9 t^2 - 1.9 t - 6.7 = 0 using the quadratic formula t = [b ± sqrt(b^2 - 4ac)] / (2a), where a = 4.9, b = -1.9, c = -6.7.

Compute the discriminant: b^2 - 4ac = (-1.9)^2 - 4(4.9)(-6.7) = 3.61 + 131.32 ≈ 134.93. Take the positive root (time cannot be negative): t = [1.9 + sqrt(134.93)] / (9.8).

Numerically, sqrt(134.93) ≈ 11.62, so t ≈ (1.9 + 11.62) / 9.8 ≈ 13.52 / 9.8 ≈ 1.38 seconds.

Rounded to one decimal place, the time is approximately 1.4 s.

If you were checking units or sanity, note that the ball rises briefly due to the initial upward velocity, then falls past the starting height and continues downward to reach the water; the total time until impact is governed by the same kinematic equation above.

MUF0121 Physics Unit 1 - Semester 2, 2025 2.14 Applications of Equations of Motion

A ball is thrown upwards from a bridge at 1.9 m s-1. How much time later does it hit the water 6.7 m below? (Give your answer to one decimal place.)

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