Answer: 2.61 s
Explanation:
We are given the following data:
[tex]s=5 ft[/tex] is the initial height of the object
[tex]v=40 ft/s[/tex] is the initial height of the object
In addition, the motion of the object is given by:
[tex]h=-16t^{2}+ vt +s[/tex] (1)
Where:
[tex]h=0 ft[/tex] is the final height of the object
[tex]t[/tex] is the time the object is in the air before hitting the ground
Rewritting (1) with the given data:
[tex]0=-16t^{2}+ 40t +5[/tex] (2)
Solving the quadratic equation with the quadratic formula [tex]t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}[/tex], where [tex]a=-16[/tex], [tex]b=40[/tex], [tex]c=5[/tex] and choosing the positive value of time:
[tex]t=\frac{-40\pm\sqrt{(-40)^{2}-4(-16)(5)}}{2(-16)}[/tex] (3)
[tex]t=2.61 s[/tex] (4) This is the time
The amount of time the shot-put spent in the air before it hits the ground is 3 sec
given the data in the question;
A person throws a shot-put from height, initial height; s = 5ft
Initial vertical velocity or speed; v = 40 ft/s
Now, it is said that, vertical motion model is;
[tex]h = -16t^2 +vt + s[/tex]
so we substitute in the given values
[tex]h = -16t^2 + (40ft/s * t) + 5ft[/tex] --------------Let this be equation 1
To calculate the amount of time the shot-put spent in the air before it hits the ground.
We say: height; h = 0 i.e when the shot-put touches the ground, the height becomes zero.
so we substitute in h=0 in equation 1
[tex]0 = -16t^2 + (40ft/s * t) + 5ft[/tex]
we rearrange
[tex]16t^2 - 40t - 5 = 0[/tex]
To solve for "t", we use the quadratic equation
[tex]x = \frac{-b\±\sqrt{b^2-4ac} }{2a} \\[/tex]
as [tex]ax^2 +bx-c = 0[/tex], we take each value from its position and substitute into the quadratic equation
so
a = 16
b = -40
c = -5
so substitute
[tex]x = \frac{-(-40)\±\sqrt{(-40)^2 - (4*16*(-5)} }{2*16} \\\\x = \frac{40\±\sqrt{1600+320} }{32}[/tex]
[tex]x = \frac{40\±\sqrt{1920} }{32}\\so\\\\x = \frac{40+\sqrt{1920} }{32} or \frac{40-\sqrt{1920} }{32} \\\\x = 2.62\\or\\ x = -0.12\\[/tex]
we know that, time can not be negative,
Hence x = 2.62 ≈ 3 sec { nearest tenth }
Therefore, the amount of time the shot-put spent in the air before it hits the ground is 3 sec
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