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yan [13]
3 years ago
13

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.
Mathematics
1 answer:
alina1380 [7]3 years ago
6 0

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

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