Question

a projectile leaves the ground with a velocity of 35 meters per second at an angle 32° what is the maximum height

Answer 1

To determine the maximum height for the projectile, we can use the timeless kinematics equation
`V _(f) ^(2) -V_(i)^(2)=2a \Delta y` and plug in what we know. For
`V _(f) ^(2)`, we know it is zero because at the highest location of flight, there is no vertical velocity. For
`V _(i) ^(2)`, we know it is
`(35sin(32))^(2)` since the vertical component is the sine of the velocity. We need to solve for
`y`, so it is left as-is. Since the only acceleration is
`g`, we can substitute -9.81 into it. Solving for the equation yields a solution of 17.55 m for
`\Delta y`.