Question

A ball was thrown straight upward with a Vi=15 m/s. How high in the air does the ball go?

Answer 1

To determine the height the ball reaches, we can use the kinematic equation for vertical motion:

$V_f^2=V_i^2+2a\cdot \Delta y$

Where:

• $V_f$ is the final velocity (which is 0 m/s at the maximum height)
• $V_i$ is the initial velocity (15 m/s)
• $a$ is the acceleration due to gravity (-9.8 m/s${}^2$)
• $\Delta y$ is the displacement or change in height

Plugging in the values we have:

$0^2=15^2+2(-9.8)\cdot \Delta y$

Simplifying the equation:

$0=225-19.6\cdot \Delta y$

Rearranging the equation to solve for $\Delta y$:

<math xmlns="http: