Question

An explosion causes debris to rise vertically with an initial velocity of 160 feet per second. What is the speed of debris when the height is 300 feet?

I am using the formula: -16t^2 + vot + h0 because the question is referring to feet.

Answer 1

Where exactly are you having trouble?

Answer 2

It always helps to setup and write out every part of the problem, starting from,\[s=-16t^2 + v_0t + s_0\]where \[v_0=160 ft/s\]\[s_0=0ft\]\[s=300\]You might begin by solving for the time, t, required to reach 300 ft.

Answer 3

Ok I solved for time and I got +/- sqrt5, is that correct?

I took the derivative of the height equation and got: -32t^2 + 160 = 0

Answer 4

I get different answers for time, check your math again or post your work on here if you want.

Answer 5

@KDsmart1985 you're derivative is basically correct except its not equal to 0. When dealing with constant acceleration equations such as these its typical the right side is another variable.
This position equation:\[s=-16t^2+v_0t+s_0\]becomes this velocity equation:\[s'=v=-32t+v_0\]and this is because the differentiation of \[s'=\frac{ ds }{ dt }\]is nothing more than a change in distance over a time interval, in other words distance/time or velocity.

Answer 6

Your times should be 2.5s and 7.5s which is logically correct. One time is the debris ascending and the other is descending. You'll get the same answer but the velocity will be negative on the descending time. I'm getting off here but so you can check your answer later, I got `v=80 ft/s`