A ball is thrown vertically upward with a velocity of 35.1m/s. Given acceleration due to gravity is - 9.8m/s2, how long will the ball be up in the air? Answer in seconds and answer to 2 decimal places (only write the numerical value, no unit)

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BlackTom AI · Accepted Answer

To determine how long the ball remains in the air, I will analyze its vertical motion under constant acceleration due to gravity.
First, identify the given values: initial upward velocity v0 = 35.1 m/s and acceleration a = -9.8 m/s^2 (negative because it acts downward).
Key concept: the time to rise to the peak occurs when the velocity becomes zero. Using v = v0 + a t, set v = 0 and solve for t: 0 = 35.1 + (-9.8) t → t_up = 35.1 / 9.8.
Compute the ascent time: 35.1 / 9.8 ≈ 3.58 s (rounded to two decimal places).
Next, because the motion is symmetric (same path up and down in constant gravity, neglecting air resistance), the time to come back to the original launch height is twice the ascent time.
Thus t_total = 2 * t_up ≈ 2 * 3.58 ≈ 7.16 s.
Another way to view it: the total time in the air equals the time to rise to the maximum height plus the time to fall back down, and for symmetrical projectile motion with an initial height of launch, those times are equal, reinforcing the doubling relationship.
Note on the units and rounding: since the instruction asks for the numerical value in seconds to two decimals, keep 7.16 as the final two-decimal figure derived from the two-step calculation above, with the intermediate step showing 3.58 s for ascent and 7.16 s for the full flight.

FFP001 系列 Threshold Quiz 3 | LA009

A ball is thrown vertically upward with a velocity of 35.1m/s. Given acceleration due to gravity is - 9.8m/s2, how long will the ball be up in the air? Answer in seconds and answer to 2 decimal places (only write the numerical value, no unit)

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