Question 966874: An object is launched straight up into the air. Its height in feet, or h, after t seconds is given by the equation h = –16t2 + 48t + 10.
1. Determine how much time has elapsed when the object is 30 feet above the ground.
2. Determine the projectile’s distance from the ground at the instant that the projectile is fired.
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! An object is launched straight up into the air. Its height in feet, or h, after t seconds is given by the equation h = –16t2 + 48t + 10.
1. Determine how much time has elapsed when the object is 30 feet above the ground.
30=-16t^2+48t+10
0=-16t^2+48t-20
4t^2-12t+5=0
(2t-5)(2t-1)=0
t=5/2=2.5
or
t=1/2
the object is 30 feet above the ground 1/2 sec after launch on the way up and 2.5 sec after launch on the way down.
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2. Determine the projectile’s distance from the ground at the instant that the projectile is fired.
h = –16t2 + 48t + 10
at the instant that the projectile is fired, t=0, so h=10=the projectile’s distance from the ground
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