SOLUTION: Height of a projectile. The height H, in feet, of a projectile with an initial velocity of 96 ft/sec is given by the equation
H=-16t squared + 96t
where t= time, in seconds. Use
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H=-16t squared + 96t
where t= time, in seconds. Use
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Question 386099: Height of a projectile. The height H, in feet, of a projectile with an initial velocity of 96 ft/sec is given by the equation
H=-16t squared + 96t
where t= time, in seconds. Use any equation-solving technique to answer the following questions.
a) How many seconds after launch is the projectile 128 ft above ground?
b) When does the projectile reach its maximum height?
c) How many seconds after launch does the projectile return to the ground?
You can put this solution on YOUR website! Height of a projectile. The height H, in feet, of a projectile with an initial velocity of 96 ft/sec is given by the equation
H=-16t squared + 96t
where t= time, in seconds. Use any equation-solving technique to answer the following questions.
a) How many seconds after launch is the projectile 128 ft above ground?
(t-4)*(t-2) = 0
t = 2 seconds (going up)
t = 4 seconds (descending)
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b) When does the projectile reach its maximum height?
Using the info above, it's at t = 3 seconds.
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c) How many seconds after launch does the projectile return to the ground?
When H = 0
H=-16t squared + 96t = 0
t = 0 (launch)
t = 6 (impact)
t in seconds
You can put this solution on YOUR website! a) 128 = -16t^2 + 96t ___ 0 = t^2 - 6t + 8 ___ there are two solutions , one going up and one coming down
b) the max height is at the vertex , which is on the axis of symmetry ___ x = -b / (2a)
in this case ___ t = -96 / [2(-16)] ___ plug the time into the original equation to find the height
