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Can anyone give me a hand with this math problem?
A person at the top of an 40 m cliff throws a rock straight up into the air with an initial
velocity of 20 m/s. The height equation is h = -4.9t
2
+ 20t + 40
a) At what times is the height of the rock 55.0 m above the ground?
b) What is rock’s maximum height?
c) How many seconds after the rock is thrown does it hit the ground?
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a. 666
b. 69
c. ask your bloody teacher
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If we answer that for you, you wouldn't learn anything.
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1. 1.89 seconds after launch
2. Max height is 58.2m
3. 7.4 seconds after launch
Of course that doesnt take into account the height of the person or when the rock actually leaves their hand.
Dont ask me for working cause I just did it quick.
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I don't know this
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One of those useless things you won't use ever again in your lifetime.
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Yet it is something that I have never forgotten.
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a) Hint: solve this equation: 55.0 = -4.9t 2 + 20t + 40 (you probably forgot a - or + after the t)
b) Hint: use your graphic calculator if you have one. Otherwise, use the derivative.
c) Hint: solve this equation: 0 = -4.9t 2 + 20t + 40
And yes, this actually IS help. I could give you the answers, it would have taken me a few minutes more, but that wouldn't be helpful since you still wouldn't know how to get those answers by yourself.
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@Danny Thanks man! exactly what I needed And @linuxthefish Dont comment if your gonna post useless shit like that man.
