SOLUTION: If a projectile is shot vertically into the air (from the ground) with an initial
velocity of 176 feet per second, its distance y (in feet) above the ground t
seconds after it is
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velocity of 176 feet per second, its distance y (in feet) above the ground t
seconds after it is
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Question 1058228: If a projectile is shot vertically into the air (from the ground) with an initial
velocity of 176 feet per second, its distance y (in feet) above the ground t
seconds after it is shot is given by y=176t+16t^2 (neglecting air resistance)
a. Find the time when y = 0 and interpret the results physically.
b. Find the times when the projectile is 16 feet off the ground. Compute answer
to two decimal places. Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website! y=0. Note, the square term is always negative.
-16t^2+176t=0
-16t(t-11)=0
t=0, t=11
t=0 is the starting point; t=11 the end point, when it hits the ground.
When the projectile is 16 feet off the ground, -16t^2+176t=16
-16t^2+176t-16=0
factor out a -16
-16(t^2-11t+1)=0
t=(1/2)(11+/- sqrt (121-4); sqrt (117)=10.82
t=(1/2)(11+/- 10.82)
t=(1/2)(21.82) or (1/2)(0.18)
t=0.09 sec and 10.91 sec (note symmetry)
