SOLUTION: The formula h=-16t^2 + v0t +s0 give sthe hieght of an object tossed upward where v0 represents the initial velocity, s0 represents the initial height, and t represents time. A golf

Question 1206216: The formula h=-16t^2 + v0t +s0 give sthe hieght of an object tossed upward where v0 represents the initial velocity, s0 represents the initial height, and t represents time. A golf ball is his straight up from the ground level with an initial velocity of 72 ft/sec.

Found 2 solutions by ikleyn, johnsss:
Answer by ikleyn(52781) (Show Source): You can put this solution on YOUR website!
.
The formula h=-16t^2 + v0t +s0 give the height of an object tossed upward where v0 represents
the initial velocity, s0 represents the initial height, and t represents time.
A golf ball is his straight up from the ground level with an initial velocity of 72 ft/sec.
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Answer by johnsss(2) (Show Source): You can put this solution on YOUR website!
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.
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