Question

9 a ball is launched vertically, from the ground, with an initial velocity of 25 m/s. when the ball reaches the ground again, what will be its velocity? -25 m/s 34.81 m/s 15.19 m/s -15.19 m/s

Explanation:

Step1: Use the kinematic equation

The kinematic equation \(v^{2}=u^{2}+2as\). Here, the initial velocity \(u = 25\ m/s\), the acceleration \(a=-g=- 9.8\ m/s^{2}\) (negative because it acts against the motion), and the displacement \(s = 0\) (since the ball returns to the ground).
Substituting into the equation: \(v^{2}=(25)^{2}+2(-9.8)(0)\)

Step2: Solve for \(v\)

Since \(s = 0\), the equation simplifies to \(v^{2}=u^{2}\). Taking the square root, \(v=\pm u\).
When the ball goes up, the velocity is positive (upward direction). When it comes back down, the direction is opposite. So \(v=-u\)

Answer:

\(-25\ m/s\)