SOLUTION: In physics, one learns that the height of an object thrown upward from an initial height of h0 feet with an initial velocity of v0 (in feet per second) is given by the formula
h(t
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Question 1120479: In physics, one learns that the height of an object thrown upward from an initial height of h0 feet with an initial velocity of v0 (in feet per second) is given by the formula
h(t)=−16t^2+v0t+h0 feet
where t is the amount of time in seconds after the ball was thrown. Also, the velocity of the object is given by
v(t)=−32t+v0. feet per second
When one uses the metric system, the equations become
h(t)=−4.9t^2+v0t+h0 meters
and
v(t)=−9.8t+v0 meters per second.
An object is projected upward from a height of 44 feet at a velocity of 44 feet per second.
After how many seconds does the object reach its maximum height?
Find the maximum height of the object.
feet
Find the velocity of the object when it hits the ground.
feet per second
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
the equation is -16t^2+44t+44
Maximum height occurs at t=-b/2a which is here -44/32 or 1.375 seconds ANSWER
This is 74.25 feet. ANSWER
Because the object was thrown at a height of 44 feet, the time in the air is not symmetric around the highest point.
Setting h=0, -16t2+44t+44=0
-4t^2+11t+11=0=4t^2-11t-11
t=(1/8)(11+/- sqrt (121+176)); sqrt (297)=17.23
t=28.23/8 or 3.53 sec as only positive root\
-32(3.53)+44=-68.96 ft/sec downward or a velocity of 68.96 ft/sec.
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