Question

An object at the top of a building with height 110 feet is thrown upward with an initial speed of 23 ft/s. Find its position z(t) above the ground t seconds after being thrown. (Use g 32 ft/sec2.)

z(t) = ___

Find the time, in seconds, it takes for the object to hit the ground.

Answer 1

Answer:

Step-by-step explanation:

Given

Height of building is $h=110\ ft$

Initial upward velocity $u=23\ ft/s$

height of object is given by

$z(t)=ut+\frac{1}{2}gt^2$

but building is already at a height of 110 ft so z(t) must be given by

$Z(t)=ut+\frac{1}{2}(-g)t^2+110$

$Z(t)=23\times t-\frac{1}{2}\times 9.8\times t^2$

$Z(t)=-16t^2+23t+110$

Time when it hits is given by when Z(t)=0[/tex]

$-16t^2+23t+110=0$

$t=\frac{-23\pm \sqrt{(23)^2-4\times (-16)\times 110}}{2\times (-16)}$

$t=\frac{-23\pm 87}{-32}$

$t=3.43\ s$

Answer 2

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 :  

                    $\rm u = 23 ft\\ h = 110 ft$  where,

• h = Height of building.

• u = initial upward velocity.

Height of object is given by,

              $\rm z(t) = ut + \dfrac{1}{2}gt^2$

But given height of building = 110 ft so, z(t) will be solved by

                   $\rm z(t) = ut + \dfrac{}{}(-g)t^2+110\\\\\rm z(t)= 23 \times t -\dfrac{1}{2} \times 9.8 \times t^2\\\\\rm z(t)=-16t^2+23t+110$

Time when the hits is given by z(t)=0[/tex]

                $\rm -16t^2=23t+110=0\\\\t=\dfrac{-23\pm\sqrt{(23)^2-4\times(-16)\times 110} }{2\times(-16)}\\\\t=\dfrac{-23\pm 87}{32}\\\\t= 3.43 s$

Therfore, its position t = 3.43s above the ground .

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