Problema Solution
if an object is thrown vertically upward with an initial velocity of v, from an original position of s,the height h at any time t is given by: h=-16t^2+vt+s (where h and s are in ft, t is in seconds and v is in ft/sec)
A package is thrown downward with an initial velocity of -5 ft/sec from a helicopter at 300ft.
a) what is the height of the package in 2 sec?
b) how long does it take to hit the ground?
round answers to 2 decimals,seperated by a comma.
Answer provided by our tutors
the distance travelled after time t with initial velocity v0 and acceleration g is
s = v0t + 1/2*gt^2 where v0 is 5 ft/s, g = 32 ft/s^2 ,and t = 2 s
s = 5*2 + 1/2*32*2^2
s = 74 ft
Since the helicopter is 300 ft above ground, the package is 300 - 74 = 226 ft above ground after 2 seconds
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set s = 300 ft in the formula and find t
300 = 5t + 16t^2
t^2 + 5/16*t - 300/16 = 0
t = - 5/32 + √(25/32^2 + 19200/32^2)
t = - 5/32 + 138.56/32
t = 4.1766 s