Question

1. A ball is dropped from a height of 225 feet. The ball’s height y (in feet) after t seconds can be modeled by y = 225-16t2 . After how many seconds does the ball hit the ground?

Answer 1

We have that
y = 225-16t²

we know that
the ball hit the ground when y=0
the point when y=0 is the x intercept of the graph
so
find the value of t for y=0
y=0
y = 225-16t²------> 0=225-16t²
16t²=225-------> t²=225/16----------> t=√(225/16) -----> t=15/4 sec
t=3.75 sec

the answer is
t=3.75 sec

see the attached figure

Answer 2

Answer:
The ball will hit the ground after 3.75 seconds

Explanation:
The equation that models the height of the ball after t seconds is given as:
y = 225 - 16t²

Now, when the ball hits the ground, the distance between the ball and the ground would be zero. This means that the height of the ball would be zero.

Therefore, to get the time at which the time would be zero, we would set the height (y) in the above equation to zero and solve for the time as follows:
0 = 225 - 16t²
16t² = 225
t² = 14.0625
either t = +√(14.0625) = 3.75 sec ..........> accepted answer
or t = -√(14.0625) = -3.75 ........> rejected as time cannot be negative

Based on the above, the ball would hit the ground after 3.75 seconds

Hope this helps :)