An arrow is shot vertically upward from a platform 45ft high at a rate of 236ft/sec. When will the arrow hit the ground? Use the formula: h=−16t2+v0t+h0.

Question

An arrow is shot vertically upward from a platform high at a rate of . When will the arrow hit the ground? Use the formula: .

Answer 1

Step One:

Set the equation equal to zero. This step represents the fact that you would like to solve for the time when the height is zero, since that is when the arrow will hit the ground.

0= -16t² + v₀t + h₀

Step Two:

Plug in 45(ft) for h0, since h₀ is the initial height.

0= -16t²+v₀t + 45(ft)

Step Three:

Plug in 236ft/sec for v₀, since v₀ represents the initial velocity.

0= -16t² + 236(ft/sec)t + 45(ft)

Step Four:

Now you have a quadratic equation and your task is to find the roots.

There are a number of ways to solve at this point. You can either use the quadratic formula or graph the equation and find when it crosses the x-axis.

i) The quadratic formula in this case yields t = (-b + - √(b^2-4ac)) / 2a

Plug in the values from the quadratic equation in step three to solve for t.

ii) Graph on a calculator or graphing program to check your answer.

Using either of these methods will yield t as approximately 15 seconds.