If an object is dropped from a 153-foot-high building, ts position (in feet abov

Question

If an object is dropped from a 153-foot-high building, ts position (in feet above the ground) is given by s(t)- -182+153, where t is the time in seconds since it was dropped. a. What is its velocity 1 second after being dropped? b. When will it hit the ground? C. What is its velocity upon impact? The object's velocity 1 second after being dropped is (Simplify your answer.) sec The object will hit the ground insec. (Round to the nearest tenth as needed.) The object's velocity upon impact is (Round to the nearest tenth as needed.) sec

Explanation / Answer

Distance traveled by object after time t is given by

s(t)= -16t2 + 153 Where s is measure from ground

a)

velocity after t= 1 second

v(t)= ds/dt= -32t

at t=1  

v(t) = -32 ft/sec

b)

when it will hit ground then s=0

thus 0= -16t2+153  

t2=153/16=9.5625 =

thus

t=3.0923 =3.1 s

c)

velocity

v(t) = ds/dt = -32 t

hitting time to ground is t= 3.1 s

thus at t =3.1 s

v(T)= -32* 3.1=-99.2 ft/sec

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