Standing on the roof of a (42.0+A) m tall building, you throw a ball straight up with an initial speed of (14.5+B) m/s. If the ball misses the building on the way down, how long will it take from you threw the ball until it lands on the ground below? Give your answer in seconds and round the answer to three significant figures. A=9, B=8
(42.0+A)= 51m
(14.5+8)= 22.5 m/s
I just need help with what equation to use..
The equation you'll use is \[\Large y = -4.9x^2 + v*x + h\] y = height of object in meters at time x x = time in seconds v = initial speed h = initial height In this case, v = 22.5 h = 51 So we'll have this equation \[\Large y = -4.9x^2 + 22.5x + 51\] after you plug in the given info
`how long will it take from you threw the ball until it lands on the ground below? ` set the height to zero then solve for x ie, replace y with 0 then isolate x
you'll use the quadratic formula
okay, hang on let me do the math.. is it okay if you check my answer after?
sure that sounds like a good plan
so this is how I set it up using the quadratic formula: \[\frac{-22.5 \pm \sqrt{22.5-4(-4.9)(51)} }{ 2(-4.9)}\]
see attached
so would then \[\frac{ -22.9 \pm \sqrt{1022.1} }{ -9.8}\] correct? or no
oh wait I just saw your message
so in the sqrt would be 1505.85?
yes `1505.85` is under the square root when you simplify
okay so what would I do next?
so we'll have this \[\Large x = \frac{-22.5 \pm \sqrt{1505.85}}{-9.8}\]
what is the square root of 1505.85 equal to?
38.8053
I think I know what to do
ok tell me what two solutions you get
I got -1.66 and 6.26 but I rounded them to the third sig fig
Ignore -1.66 why? because x = -1.66 makes no sense. Recall that x is the time in seconds. We can't have a negative time value
So the only solution is x = 6.26
oh okay, thank you!
you're welcome
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