Answer :

Answer:

Ball will take 1.9591 sec to reach the ground

Explanation:

We have given initial velocity of the ball u = 10 m/sec

Height from which ball is thrown h = 38.4 m

Acceleration due to gravity [tex]g=9.8m/sec^2[/tex]

From second equation of motion we know that [tex]h=ut+\frac{1}{2}gt^2[/tex]

[tex]38.4=10t+4.9t^2[/tex]

[tex]4.9t^2+10t-38.4=0[/tex]

After solving equation for t

t= -4 sec and t = 1.9591 sec

As time can not be negative so time will be 1.9591 sec

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Rewritten by : Barada

Final answer:

To find the time it takes for a ball thrown straight down at 10.0 m/s from a 38.4 m-high building to reach the ground, we use the kinematic equations of motion, resulting in a time of approximately 2.29 seconds.

Explanation:

Calculating Free Fall Time of a Ball

A student asks about the time it takes for a ball thrown straight down at 10.0 m/s from a 38.4 m-high building to reach the ground. To solve this, we use the kinematic equations of motion. The formula to determine the time (t) is:

h = v_i * t + (1/2) * g * t²

where h is the height (38.4 m), v_i is the initial velocity (10.0 m/s downward), and g is the acceleration due to gravity (approximately 9.81 m/s² downward).

Plugging the known values into the equation gives us:

38.4 = 10 * t + (1/2) * 9.81 * t²

Rearranging the equation to fit the quadratic equation format (ax^2 + bx + c = 0), we get:

(4.905)t² + (10)t - 38.4 = 0

Using the quadratic formula, t = (-b ± √(b² - 4ac))/(2a), we find two possible values for t, but only the positive value is physically meaningful. Eventually, we find:

t ≈ 2.29 seconds (rounded to two decimal places).

So the ball will take approximately 2.29 seconds to hit the ground.