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

Answer:

t = 0.118 seconds

Explanation:

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

$y=v_0t-\frac{1}{2}gt^2$

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,

$y=17.6t^{}-\frac{1}{2}\times9.8t^2$

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

$2=17.6t-\frac{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\times(-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.