Do I use substitution then quadratic formula for solving 1_d projectile problems?
quick sample
A stone if dropped from a cliff. It takes 3.0s for the sound of the splash of stone to reach the top of the top of cliff.
I use substitution then quadratic formula right?
Thanks
quick sample
A stone if dropped from a cliff. It takes 3.0s for the sound of the splash of stone to reach the top of the top of cliff.
I use substitution then quadratic formula right?
Thanks
-
If it is dropped then vi = 0. That makes your problem a little simpler. The question is does the time it takes the sound to get back to you matter or not? If it doesn't, then the formula is
d = 1/2 at^2 and you are likely trying to find d.
If it does matter then the problem is a good deal more complicated. Now what you have is
d = 1/2 a t1^2
d = 345 m/s * t2
t1 + t2 = 3 seconds. 345 m/s is the speed of sound in air. It varies depending on where you look for the constant.
This would give you a quadratic
345 (3 - t1) = 4.9 t^2 remove the brackets.
1035 - 345 t2 = 4.9 t^2
4.9 t^2 + 345 t2 - 1035 = 0
a = 4.9
b = 345
c = - 1035
t = 2.88 there is another value, but it is minus and has an absolute value much larger than 3.
So the time it takes to go down is 2.88 seconds. The time to come back is 0.02 seconds. I have a feeling that the second condition is much more complicated than you need. However this will give you the height of the cliff
a. if you ignore the speed of sound d = 4.9 * 3^2 = 44.1 meters.
b. if you need the speed of sound then the height is d = 4.9 * 2.88^2 = 40.64
d = 1/2 at^2 and you are likely trying to find d.
If it does matter then the problem is a good deal more complicated. Now what you have is
d = 1/2 a t1^2
d = 345 m/s * t2
t1 + t2 = 3 seconds. 345 m/s is the speed of sound in air. It varies depending on where you look for the constant.
This would give you a quadratic
345 (3 - t1) = 4.9 t^2 remove the brackets.
1035 - 345 t2 = 4.9 t^2
4.9 t^2 + 345 t2 - 1035 = 0
a = 4.9
b = 345
c = - 1035
t = 2.88 there is another value, but it is minus and has an absolute value much larger than 3.
So the time it takes to go down is 2.88 seconds. The time to come back is 0.02 seconds. I have a feeling that the second condition is much more complicated than you need. However this will give you the height of the cliff
a. if you ignore the speed of sound d = 4.9 * 3^2 = 44.1 meters.
b. if you need the speed of sound then the height is d = 4.9 * 2.88^2 = 40.64