Asked by hala aly
A golf ball is hit from the top of a tee. The quadratic equation
𝑦 = −5𝑥2 + 20𝑥 + 0.05 describes its height, 𝑦, in metres as time, 𝑥, in seconds
passes. Determine how long the ball is in the air
𝑦 = −5𝑥2 + 20𝑥 + 0.05 describes its height, 𝑦, in metres as time, 𝑥, in seconds
passes. Determine how long the ball is in the air
Answers
Answered by
Damon
Well, I would write that as
y = 0.05 + 20 t - 5 t^2
the initial height when t = 0 is y = 0.05
the initial velocity up is 20
and the acceleration down is 2 * 5 = 10
so we usually write that as
h = Hi + Vi t - (1/2) g t^2
that said returning to the problem
5 t^2 - 20 t - 0.05 = 0
what is t when h = 0 standard old quadratic
t = [ 20 +/- sqrt (400 + 1) ] / 10
looks like about 4 seconds
By the way. 0.05 meters is a very low tree !! Typo?
y = 0.05 + 20 t - 5 t^2
the initial height when t = 0 is y = 0.05
the initial velocity up is 20
and the acceleration down is 2 * 5 = 10
so we usually write that as
h = Hi + Vi t - (1/2) g t^2
that said returning to the problem
5 t^2 - 20 t - 0.05 = 0
what is t when h = 0 standard old quadratic
t = [ 20 +/- sqrt (400 + 1) ] / 10
looks like about 4 seconds
By the way. 0.05 meters is a very low tree !! Typo?
Answered by
Damon
Oh, sorry, misread tree for tee :)
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