Problema Solution
An object is thrown straight up from the top of a 50ft tower. Its initial velocity is 120ft/sec. The equation that models the height h of the object above the ground, t seconds after it was thrown is:
h(t)=-16t^2+120t+50
a.) How long after it is thrown until it reaches it's maximum height?
b.) What is its maximum height?
c.) How high after 1.5 seconds?
d.) After how many seconds does it hit the ground?
e.) At what time(s) is it 175ft above the ground?
Answer provided by our tutors
a.) How long after it is thrown until it reaches it's maximum height?
we need to find the maximum of the function h(t)=-16t^2+120t+50
since the quotient in front of t^2 is -1<0 the function has maximum in its vertex:
t = -b/2a
t = -120/(2*(-16))
t = 3.75 s
click here to see the step by step calculation:
b.) What is its maximum height?
h max = c - b^2/(4a), where a =-16, b=120, c=50
h max = 50 - 120^2/(4*(-16))
h max = 275 ft
the maximum height is 275 feet.
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c.) How high after 1.5 seconds?
t = 1.5 s plug into h(t)=-16t^2+120t+50
h(1.5)=-16*1.5^2+120*1.5+50
h(1.5) = 194 ft
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d.) After how many seconds does it hit the ground?
the object hits the ground when h(t) = 0 for some t>0
-16t^2+120t+50 = 0
by solving we find:
t = 7.9 s
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e.) At what time(s) is it 175ft above the ground?
we need to find t, t>0 such that h(t)= = 175 that is
-16t^2+120t+50 = 175
by solving we find:
t1 = 1.25 s
t2 = 6.25 s
click here to see the step by step solution of the equation:
the object will be 175 ft above the ground 1.25 seconds and 6.25 second after it was thrown straight up.