Question

a ball is thrown upward in the air with an initial velocity of 40 m/s. how long does it take to reach the top of its path? (use g = 10m/s²)

Explanation:

Step1: Identify the kinematic - equation

We use the equation $v = v_0+at$. At the top of the path, $v = 0$. The initial velocity $v_0 = 40$ m/s and the acceleration $a=-g=- 10$ m/s² (negative because it acts against the motion).

Step2: Solve for time $t$

From $v = v_0+at$, we can re - arrange it to $t=\frac{v - v_0}{a}$. Substituting $v = 0$, $v_0 = 40$ m/s and $a=-10$ m/s², we get $t=\frac{0 - 40}{-10}$.
$t=\frac{-40}{-10}=4$ s

Answer:

D. 4 seconds