Question

A projectile is shot at a velocity of `2 m/s` and an angle of `pi/4`. What is the projectile's peak height?

Answer 1

`y_"max"(0.14)~~0.10" meters"` (that's about 4 inches)

Explanation:

You have a projectile and you want to know the vertical position (i.e., the `y`-value) at the projectile's peak. The general equation is

`y=(v_0sin(theta_0))t-1/2 g t^2`

You are given

• `v_0=2` m/s (pretty slow, about 4.5 mph)
• `theta_0=pi"/"4`
• `g=9.8` m/s`""^2`

Plugging in those values we get `y(t)`

`y(t) = (2" m/s")sin(pi"/"4)t-0.5(9.8" m/s"^2)t^2`
`" "=(sqrt(2))t-4.9t^2` meters

A graphing calculator will find your max height at `y~~0.10`
graph{y=2sin(pi/4)x-0.5(9.8)x^2[-.01,.3,-.01,.11]}

The time at which the this function is maximum is discovered by solving for `t` when `dy"/"dt=0`.

`d/dt(y)=y'(t)=sqrt(2)-9.8t`

Set `y'(t)=0`

`sqrt(2)-9.8t=0`

Now solve for `t`
`9.8t=sqrt(2)`
`t=sqrt(2)"/"9.8`
`t~~0.14` seconds

So the maximum height occurs at that time

`y_"max"(0.14)~~(2"m/s")sin(pi"/"4)(0.14"s")-1"/"2(9.8"m/s")(0.14"s")^2`
`y_"max"(0.14)~~0.10" meters"` (that's about 4 inches)