Asked by Olivia N J
Suppose this firework is set to explode 3 seconds after it is launched. At what height will this firework be when it explodes?
My equation is h(t)=-16t^2+120t+20
How do I do this?
My equation is h(t)=-16t^2+120t+20
How do I do this?
Answers
Answered by
Damon
h = 20 + 120 * 3 - 16 (9)
= 20 + 360 - 144
in feet where g = 32 ft/s^2
= 20 + 360 - 144
in feet where g = 32 ft/s^2
Answered by
Damon
this is because in feet and seconds
acceleration of gravity, g = 32 ft/s^2
then velocity v = Vi - g t
where Vi is initial velocity and g = 32 ft/s^2
and height h is
h = Hi + Vi t - (1/2)g t^2
in this problem
h = Hi + Vi t - 16 t^2
where Hi = 20 and Vi = 120
acceleration of gravity, g = 32 ft/s^2
then velocity v = Vi - g t
where Vi is initial velocity and g = 32 ft/s^2
and height h is
h = Hi + Vi t - (1/2)g t^2
in this problem
h = Hi + Vi t - 16 t^2
where Hi = 20 and Vi = 120
Answered by
Anonymous
Thanks Damon!
Answered by
Damon
You are welcome.
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