Question

A ball is thrown straight up into the air, with an initial speed of 18.2 m/s. How far has it gone after 1.00 s?

Answer 1

It's vertical position in the air at time
`t` is

`y=\left(18.2\,(\mathrm m)/(\mathrm s)\right)t-\frac g2t^2`

where
`g=9.80\,(\mathrm m)/(\mathrm s^2)` is the acceleration due to gravity. After 1.00 second has passed, its position is

`\left(18.2\,(\mathrm m)/(\mathrm s)\right)(1.00\,\mathrm s)-\frac g2(1.00\,\mathrm s)^2=13.3\,\mathrm m`

Answer 2

The ball is going to reach 13.3 meters after 1 second.

Step-by-step explanation:

The vertical movement of the ball can be described by the kinematic equations of y component for the position:

`y=yo+vy_(o)t+1/2*ayt^2`

Where yo is the initial position and depends on where the system of reference is located. For this purpose is useful to consider the origin of the system of reference at the ground level, thus yo will be 0. The vyo coefficient is the initial vertical velocity and is given as 18.2 m/s. The ay coefficient is the vertical acceleration and due there is only the ball weight acting, the acceleration is taken of -9.8 m/s^2 (acceleration of gravity ). The time t must be expressed in seconds to obtaining the position in meters.

`y=18.2(m)/(s)*t+1/2*(-9.8(m)/(s^2))*t^2`

Evaluating the equation for a time of 1 second:

`y=18.2(m)/(s)*(1 s)+1/2*(-9.8(m)/(s^2))* (1 s)^2`

`y=18.2 m+ (-4.9 m)`

`y=13.3 m`