Answer to Question #145756 in Algebra for liam donohue

Question

You toss a rock up vertically at an initial speed of 40 feet per second and release it at an initial height of 6 feet. The rock will remain in the air for how many
seconds.
It will reach a maximum height of how many
feet after how many
seconds.
Note: Ignore air resistance.

Answer 1

Let`s start with recognising given valuables:

V₀=40 feet/second,

h₀=6 feet

And we need to find:

t_total=?, h_max=?, t₁=?

t₁ is time which is spent to reach maximum height.

Firstly, let`s start to compute with the maximum height:

h_max=h₀+V₀²/(2*g) --> g=9.81 m/s² -->g=32.18 feet/second

we have everything to calculate this and the result is:

h_max=6+40²/(2*32.18)=30.86 feet.

Secondly we calculate time to reach maximum height:

t₁=$\sqrt{\smash[b]{(2*(hmax-h0)/g)}}$ =1.24 second

Lastly, we compute the value of total time of flying:

t_total=t₁+$\sqrt{\smash[b]{(2*hmax/g)}}$ =2.62 second.

We found everything which are asked.