Question 1133975: At a fireworks stand show, a 3 in shell is shot from a mortar at an angle of 75 degree. The height, y (in feet) , of the shell t sec after being shot from the mortar is given by the quadratic equation
Y= - 16t^2 + 144 and the horizontal distance of the shell from the mortar, x ( in feet), is given by the linear equation x=39t.
A) How high is the shell after 3 seconds?
B) What is the shell’s horizontal distance from the mortar after 3 seconds?
C) The maximum height is reached when the shell explodes. How high is the shell when it burst after 4.5 seconds?
D) What is the shell’s horizontal distance from launching point when it explodes? ( Round to the nearest foot).
Answer by Glaviolette(140)   (Show Source):
You can put this solution on YOUR website! I'm a bit confused about the results from the equation for height that was given. However...here is what I've come up with.
a. t = 3 sec
y = -16(3)^2 + 144
y = 0 ft (The shell hits the ground at 3 seconds)
b. t = 3 sec
x = 39(3)
x = 117 ft
c. t = 4.5 sec (I don't agree that this is when it reaches it's maximum height since it hit the ground at 3 sec and the answer to this is negative)
y = -16(4.5)^2 + 144 = -180 ft
d. x = 39(4.5) = 175.5 ft (although this doesn't make sense based on previous answers)
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