A.) y = ut - 1/2gt^2 + 12
y = 6t - 1/2(9.8)t^2 + 12
y = 6t - 4.9t^2 + 12
b.) The y-intercept is 12
c.) The y-intercept represent the height of the ball at time t = 0
d.) At the time the ball reaches the ground, y = 0
-4.9t^2 + 6t + 12 = 0
t = 2.29 seconds
Therefore, it took the ball 2.29 seconds to reach the ground.
I think it’s between 100-200 like 150 cause it’s near a 100 and more then a 100 too.
X
3
+
3
x
2
y
+
2
x
y
2
+
y
3
+
x
2
+
3
x
y
+
y
2
