Question

A ball is thrown upwards into the air with an initial velocity of 11.83 m/s. What is the maximum height that the ball traveled during that time (assuming there is no air resistance).

Answer 1

7.11 meters

Step-by-step explanation

to solve this we need to use this formula:

`y_(\max )=((v_f-v_(1)^2))/(2g)`

Step 1

Let

`\begin{gathered} v_f=0\text{ ( when it reaches the ma}\Xi mun\text{ heigth)} \\ v_0=\text{ initial sp}eed=11.83\text{ m/s} \\ g=9.8\text{ }\frac{\text{m}}{s^2} \end{gathered}`

now,replace and calculate

`\begin{gathered} y_(\max )=((v_f-v_(1)^2))/(2g) \\ y_(\max )=((-11.83(m)/(s))^2)/(2\cdot9.8(m)/(s^2)) \\ y_(\max )=(139.4989)/(19.6)m \\ y_(\max )=7.11\text{ m} \end{gathered}`

therefore, the answer is

7.11 meters

I hope this helps you