Respuesta :
Answer:
Step-by-step explanation:
Given
Height of building is [tex]h=110\ ft[/tex]
Initial upward velocity [tex]u=23\ ft/s[/tex]
height of object is given by
[tex]z(t)=ut+\frac{1}{2}gt^2[/tex]
but building is already at a height of 110 ft so z(t) must be given by
[tex]Z(t)=ut+\frac{1}{2}(-g)t^2+110[/tex]
[tex]Z(t)=23\times t-\frac{1}{2}\times 9.8\times t^2[/tex]
[tex]Z(t)=-16t^2+23t+110[/tex]
Time when it hits is given by when Z(t)=0[/tex]
[tex]-16t^2+23t+110=0[/tex]
[tex]t=\frac{-23\pm \sqrt{(23)^2-4\times (-16)\times 110}}{2\times (-16)}[/tex]
[tex]t=\frac{-23\pm 87}{-32}[/tex]
[tex]t=3.43\ s[/tex]
An object at the top of a building with height 110 feet is thrown upward with an initial speed of 23 ft/s, its position
t = 3.43s above the ground.
Given :
[tex]\rm u = 23 ft\\ h = 110 ft[/tex] where,
- h = Height of building.
- u = initial upward velocity.
Height of object is given by,
[tex]\rm z(t) = ut + \dfrac{1}{2}gt^2[/tex]
But given height of building = 110 ft so, z(t) will be solved by
[tex]\rm z(t) = ut + \dfrac{}{}(-g)t^2+110\\\\\rm z(t)= 23 \times t -\dfrac{1}{2} \times 9.8 \times t^2\\\\\rm z(t)=-16t^2+23t+110\\[/tex]
Time when the hits is given by z(t)=0[/tex]
[tex]\rm -16t^2=23t+110=0\\\\t=\dfrac{-23\pm\sqrt{(23)^2-4\times(-16)\times 110} }{2\times(-16)}\\\\t=\dfrac{-23\pm 87}{32}\\\\t= 3.43 s[/tex]
Therfore, its position t = 3.43s above the ground .
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