Question

A ball is hit straight up in the air with an initial velocity of 17.6m/a. At what time does it cross the height of 2 m going up?

Answer 1

t = 0.118 seconds

Explanation:

The vertical distance in a straight-up motion of the ball is described by

`y=v_0t-(1)/(2)gt^2`

where

v0 = vertical velocity of the ball

g = acceleration due to gravity

t = time after the launch

Now in our case, v_0 = 17.6 m/s and g= 9.8 m/s^2; therefore,

`y=17.6t^{}-(1)/(2)*9.8t^2`

When the ball crosses the height of 2m, we have y = 2; therefore, the above equation gives

`2=17.6t-(9.8)/(2)t^2`
`\Rightarrow2=17.6t-4.9t^2`

subtracting 2 from both sides gives

`17.6t-4.9t^2-2=0`

The above can be rewritten in a more familiar form by a bit of rearranging of the terms.

`-4.9t^2+17.6t-2=0`

This is a quadratic equation, and therefore, the values of t can be found using the quadratic formula.

Using the quadratic formula, we get two possible values of t:

`t=\frac{-17.6\pm\sqrt[]{17.6^2-4(-4.9)(-2)}}{2*(-4.9)}`
`\begin{gathered} \Rightarrow t=0.1175 \\ t=3.474 \end{gathered}`

Which value of t is correct?

The two values of t tell us that the ball crosses the y = 2 checkpoint twice. First when going up and second when coming down. Since we want the time when the ball is going up, we choose t= 0.1175 (because the ball goes up first and then it comes down).

Hence, after 0.118s (rounded to the nearest thousandth), the ball crosses the height of 2m going up.