Final answer:
The height above the ground when the ball was thrown is approximately 158.75 m. The velocity of the ball just before it hits the ground is -20 m/s.
Step-by-step explanation:
To solve this problem, we can start by breaking it down into two parts: the vertical motion and the horizontal motion. For the vertical motion, we can use the kinematic equation:
h = h0 + v0t + 0.5gt^2
Where h is the height above the ground, h0 is the initial height (which is the same as the height of the ground), v0 is the initial vertical velocity (in this case, 20 m/s), t is the time (15.5 s), and g is the acceleration due to gravity (approximately -9.8 m/s^2).
Plugging in the values, we get:
h = 0 + 20 * 15.5 + 0.5 * -9.8 * (15.5)^2
Simplifying the equation, we find that the height above the ground when the ball was thrown is approximately 158.75 m.
For the horizontal motion, we know that the horizontal velocity remains constant throughout the motion. Therefore, the velocity of the ball just before it hits the ground is the same as its initial horizontal velocity. So the answer is -20 m/s.