Question 1206216
To solve the problem using the equation. 
Source: https://mipromedio.co/blog/como-calcular-promedio-universitario/
ℎ
=
−
16
𝑡
2
+
𝑣
0
𝑡
+
𝑠
0
h=−16t 
2
 +v 
0
​
 t+s 
0
​
 , let's break it down:

Given:

𝑣
0
=
72
v 
0
​
 =72 ft/sec (initial velocity)

𝑠
0
=
0
s 
0
​
 =0 (since the ball is hit from ground level)

The equation becomes 
ℎ
=
−
16
𝑡
2
+
72
𝑡
h=−16t 
2
 +72t

This formula describes the height of the golf ball at any given time 
𝑡
t. If you need to find specific values, like the maximum height or when the ball hits the ground, you'd follow these steps:

Maximum Height: The maximum height occurs at the vertex of the parabola, which can be found by the formula 
𝑡
max
=
−
𝑣
0
2
𝑎
t 
max
​
 =− 
2a
v 
0
​
 
​
 . For this equation, 
𝑎
=
−
16
a=−16 and 
𝑣
0
=
72
v 
0
​
 =72, so:

𝑡
max
=
−
72
2
(
−
16
)
=
72
32
=
2.25
 
seconds
t 
max
​
 = 
2(−16)
−72
​
 = 
32
72
​
 =2.25seconds
Then substitute this into the equation to find the height at this time:

ℎ
=
−
16
(
2.25
)
2
+
72
(
2.25
)
=
−
16
(
5.0625
)
+
162
=
−
81
+
162
=
81
 
ft
h=−16(2.25) 
2
 +72(2.25)=−16(5.0625)+162=−81+162=81ft
So, the maximum height is 81 feet.

Time to hit the ground: The ball will hit the ground when 
ℎ
=
0
h=0. Set the height equation equal to zero and solve for 
𝑡
t:

0
=
−
16
𝑡
2
+
72
𝑡
0=−16t 
2
 +72t
Factor:

0
=
𝑡
(
−
16
𝑡
+
72
)
0=t(−16t+72)
So, 
𝑡
=
0
t=0 (at launch) or 
𝑡
=
72
16
=
4.5
 
seconds
t= 
16
72
​
 =4.5seconds.