Question

A ball is thrown from a height of 3 meters with an initial downward velocity of 5 m/s The ball's height h (in meters) after t seconds is given by the following. how long after the ball is thrown does it hit the ground

Answer 1

The ball hits the ground at a time of 1.433 seconds.

How to calculate the time until the ball hits the ground

In this problem we have the case of a ball that experiments a free fall, that is, an uniform accelerated motion due to gravity. The position of the ball in time is described by a quadratic equation:

y = y' + v' · t + 0.5 · g · t²

Where:

• y' - Initial position, in meters.
• v' - Initial speed, in meters per second.
• t - Time, in seconds.
• g - Gravitational acceleration, in meters per square second.
• y - Final position, in meters.

If we know that y' = 3 m, v' = 5 m / s, g = - 9.807 m / s², y = 0 m, then the time of the ball until it hits the ground is:

0 = 3 + 5 · t - 4.904 · t²

Then, by the quadratic formula:

t = - v' / g ± (1 / g) · √[v'² - 2 · g · y']

t = - 5 / (- 9.807) ± [1 / (- 9.807)] · [5² - 2 · (- 9.807) · 3]

t₁ = 1.433 s. or t₂ = - 0.424 s.

The hitting time of the ball is 1.433 seconds.

To learn more on free fall: https://brainly.com/question/13796105

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