A ball is thrown vertically in the air with a velocity of 90ft/s. Use the projectile formula h=−16t2+v0t to determine at what time(s), in seconds, the ball is at a height of 120ft. Round your answer(s) to the nearest tenth of a second.
Answers
Answer:
The ball is at a height of 120ft when t = 2.2 and when t = 3.5.
Step-by-step explanation:
h = -16t^{2} + v(0)th=−16t
2
+v(0)t
A ball is thrown vertically in the air with a velocity of 90ft/s.
This means that v(0) = 90. So
h = -16t^{2} + 90th=−16t
2
+90t
Use the projectile formula h=−16t2+v0t to determine at what time(s), in seconds, the ball is at a height of 120ft.
This is t when
-16t^{2} + 90t = 120−16t
2
+90t=120
16t^{2} - 90t + 120 = 016t
2
−90t+120=0
Finding t
Given a second order polynomial expressed by the following equation:
ax^{2} + bx + c, a\neq0ax
2
+bx+c,a
=0 .
This polynomial has roots x_{1}, x_{2}x
1
,x
2
such that ax^{2} + bx + c = (x - x_{1})*(x - x_{2})ax
2
+bx+c=(x−x
1
)∗(x−x
2
) , given by the following formulas:
x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}x
1
=
2∗a
−b+
△
x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}x
2
=
2∗a
−b−
△
\bigtriangleup = b^{2} - 4ac△=b
2
−4ac
In this problem:
16t^{2} - 90t + 120 = 016t
2
−90t+120=0
So a = 16, b = -90, c = 120a=16,b=−90,c=120
\bigtriangleup = (-90)^{2} - 4*16*120 = 420△=(−90)
2
−4∗16∗120=420
t_{1} = \frac{-(-90) + \sqrt{420}}{2*16} = 3.5t
1
=
2∗16
−(−90)+
420
=3.5
t_{2} = \frac{-(-90) - \sqrt{420}}{2*16} = 2.2t
2
=
2∗16
−(−90)−
420
=2.2
The ball is at a height of 120ft when t = 2.2 and when t = 3.5.
Answer:
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Step-by-step explanation: