a ball is thrown upward at 96 feet per second from the top of a 100 foot building.

1.when will the ball obtain maximum height

2. when will the ball hit the ground?

s = 100 + 96t - 16t^2

ds/dt = 96 - 32t
ds/dt = 0 when t = 3

s = 0 when

100 + 96t - 16t^2 = 0
when t = 6.9

To answer these questions, we need to use the equations of motion. Let's break it down step by step.

1. When will the ball obtain maximum height?
To find the time at which the ball reaches its maximum height, we need to use the equation of motion for vertical motion:

h = ut - (1/2)gt^2

Where:
- h is the height
- u is the initial vertical velocity (96 ft/s, as given)
- g is the acceleration due to gravity (32 ft/s^2)

At the maximum height, the velocity of the ball becomes zero. Therefore, we can set the final velocity (v) to zero in the equation:

0 = 96t - 16t^2

Simplifying the equation:

16t^2 - 96t = 0

Factoring out t:

t(16t - 96) = 0

This equation has two solutions:
t = 0 (initial time) and t = 6 seconds

However, since we're interested in the time it takes for the ball to reach its maximum height, we disregard t = 0. Therefore, the ball will reach its maximum height after 6 seconds.

2. When will the ball hit the ground?
To find when the ball hits the ground, we need to determine the time it takes for the ball to reach a height of 0 feet.

Using the same equation of motion:

h = ut - (1/2)gt^2

Setting h = 0:

0 = 96t - 16t^2

Simplifying:

16t^2 - 96t = 0

Factoring out t:

t(16t-96) = 0

This equation also has two solutions:
t = 0 (initial time) and t = 6 seconds

Since t = 0 represents the starting point, we disregard it. Therefore, the ball will hit the ground after 6 seconds.

To summarize:
1. The ball will reach its maximum height after 6 seconds.
2. The ball will hit the ground after 6 seconds.