To find the time it takes for the ball to hit the ground, we can use the equation of motion for vertical motion under constant acceleration. In this case, the acceleration is due to gravity, which is approximately 9.8 m/s² near the surface of the Earth.
The equation for the displacement of an object undergoing vertical motion is given by:
s = ut + (1/2)at²,
where: s is the displacement, u is the initial velocity, t is the time, a is the acceleration.
In this case, the ball is thrown upwards, so the initial velocity (u) is 10 m/s (considered positive as it is in the upward direction), the displacement (s) is -20 m (considered negative as it is downward), and the acceleration (a) is -9.8 m/s² (negative as it acts in the opposite direction to the initial velocity).
Plugging in these values into the equation, we have:
-20 = 10t - (1/2)(9.8)t².
Simplifying and rearranging the equation, we get:
(1/2)(9.8)t² - 10t - 20 = 0.
This is a quadratic equation in terms of t. Solving this equation, we find two possible solutions: t ≈ 0.85 s and t ≈ 2.35 s. Since we're interested in the time it takes for the ball to hit the ground, we discard the smaller positive value. Therefore, the time it takes for the ball to hit the ground is approximately 2.35 seconds.