Question

if a ball is thrown straight up with a velocity of 30 m/s, how long until it can be caught at the same height? (assume no air resistance)

Explanation:

Step1: Identify the kinematic - equation

We use the displacement formula $y - y_0=v_0t-\frac{1}{2}gt^2$. Since the ball is caught at the same height, $y - y_0 = 0$. The initial velocity $v_0 = 30\ m/s$ and $g = 10\ m/s^2$.

Step2: Solve the equation for time

Set $y - y_0 = 0$, so $0=v_0t-\frac{1}{2}gt^2$. Factor out $t$: $t(v_0-\frac{1}{2}gt)=0$. One solution is $t = 0$ (corresponds to the initial - throw time). For the non - zero solution, we solve $v_0-\frac{1}{2}gt=0$. Rearranging gives $t=\frac{2v_0}{g}$.

Step3: Substitute the values

Substitute $v_0 = 30\ m/s$ and $g = 10\ m/s^2$ into $t=\frac{2v_0}{g}$. Then $t=\frac{2\times30}{10}=6\ s$.

Answer:

6 s