$a_1=96;\ a_2=48;\ a_3=24;\ a_4=12;\ a_5=6\\\\a_1=a_2:2\to96:2=48\\a_2=a_3:2\to48:2=24\\a_3=a_2:2\to24:2=12\\a_4=a_3:2\to12:2=6\\\\therefore\\\\a_n=96:2^{n-1}=\frac {96}{2^{n-1}}\\\\check:\\a_1=\frac{96}{2^{1-1}}=\frac{96}{1^0}=96\\a_2=\frac{96}{2^{2-1}}=\frac{96}{2^1}=\frac{96}{2}=48\\a_3=\frac{96}{2^{3-1}}=\frac{96}{2^2}=\frac{96}{4}=24\\\vdots$

8 0

3 years ago

In a ballistic pendulum experiment, projectile 1 results in a max height,h, of the pendulum equal to 2.6cm. A second projectile

prohojiy [21]

The second projectile was travelling 1.4 times faster than the first. I am hoping that this answer has satisfied your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.

4 0

2 years ago