To find the time it takes for the ball to hit the ground, we can use the kinematic equation for vertical motion:
h = ut + (1/2)gt^2
where h is the initial height (20 m), u is the initial velocity (10 m/s), g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time.
Since the ball is thrown upward, its initial velocity is positive (+10 m/s), and the acceleration due to gravity is acting downward, so we take g as a negative value (-9.8 m/s^2).
At the highest point of its trajectory, the ball will momentarily stop, and its final velocity will be zero. Therefore, we can use this information to find the time it takes to reach the highest point:
0 = u + gt 0 = 10 - 9.8t 9.8t = 10 t = 10 / 9.8 t ≈ 1.02 seconds
Now, let's calculate the time it takes for the ball to hit the ground by using the initial height (h = 20 m) and solving for time:
h = ut + (1/2)gt^2 20 = 10t + (1/2)(-9.8)t^2 0 = -4.9t^2 + 10t - 20
We can solve this quadratic equation using the quadratic formula:
t = (-b ± sqrt(b^2 - 4ac)) / (2a)
For this equation, a = -4.9, b = 10, and c = -20. Substituting these values into the quadratic formula:
t = (-10 ± sqrt(10^2 - 4(-4.9)(-20))) / (2(-4.9))
Simplifying further:
t = (-10 ± sqrt(100 - 392)) / (-9.8) t = (-10 ± sqrt(-292)) / (-9.8)
Since the square root of a negative number is not a real number, the ball will not hit the ground again. Instead, it will reach a maximum height of 20 meters and then fall back down.
Therefore, the time it takes for the ball to hit the ground is not applicable in this scenario.