Your answer would be be subtract 3
hope it helps
Answer:
The maximum height that the ball reaches is sf = 31 feet from the ground
Step-by-step explanation:
Solution:-
- A ball is thrown vertically upwards from an initial elevation of si = 6 feet from the ground.
- The velocity with which the ball was thrown up, vi = 40 ft/s
- We can determine the maximum height of the ball when it is thrown vertically up by using the 3rd kinematic equation of motion.
vf^2 - vi^2 = 2*g*( sf - si )
Where,
vf : The final velocity of the ball, for maximum height it is = 0
sf : The final height of the ball from the ground
g : Gravitational constant = -32 ft/s^2
0 - 40^2 = -2*32*( sf - 6 )
sf - 6 = 25
sf = 31 feet
- The maximum height that the ball reaches is sf = 31 feet from the ground.
B. Explain the steps to solve the absolute value equation. What do the solutions represent?