Question

If a ball is thrown straight up in the air and catch it as it comes down. The recorded time of travel for the ball from start till it got caught is 0.53sec. What is the initial velocity and how high does the ball go?

Answer 1

The initial velocity is -5.2 m/s

The height is 1.38 m

Explanation:

We have to calculate the velocity, we must follow the following formula of the vertical movement upwards:

`\begin{gathered} v_i=-g\cdot t \\ \text{where,} \\ g=\text{gravity = 9.8 }(m)/(s^2) \\ t=\text{time = 0.53 sec} \end{gathered}`

replacing:

`\begin{gathered} v_i=-9.8\cdot0.53 \\ v_i=-5.2\text{ m/s} \end{gathered}`

Now, to calculate the height that the ball reached, we use the following formula:

`\begin{gathered} h=(-v^2_i)/(2g) \\ \text{ replacing} \\ h=(-\mleft(-5.2\mright)^2)/(2\cdot9.8) \\ h=1.38\text{ m} \end{gathered}`