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 $\Delta y$.