Question

A projectile is shot at a velocity of `1 m/s` and an angle of `pi/4`. What is the projectile's peak height?

Answer 1

If the projectile is shot at a velocity of `1` m/s then the vertical component of velocity is given by `1 * sin(pi/4) = 1/sqrt(2)` m/s and the horizontal component of velocity is given by `1 * cos(pi/4) = 1/sqrt(2)` m/s.

The maximum height will occur when the vertical component of velocity equals `0`.

We now need to find a kinematic equation that will make sense for this problem.

`v_y^2 = v_(y_0)^2 + 2a_y(Deltay)`

We will be solving for `Delta y`, the maximum height of the projectile. `a_y` will be the acceleration due to gravity, `9.8" "m/s^2`.

`0 = 1/2 + 2(-9.8)Deltay`

`Delta y = 0.026` meters, or `2.6` cm.

Hopefully this helps!