Respuesta :
Answer:
t = 0.118 seconds
Explanation:
The vertical distance in a straight-up motion of the ball is described by
[tex]y=v_0t-\frac{1}{2}gt^2[/tex]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,
[tex]y=17.6t^{}-\frac{1}{2}\times9.8t^2[/tex]When the ball crosses the height of 2m, we have y = 2; therefore, the above equation gives
[tex]2=17.6t-\frac{9.8}{2}t^2[/tex][tex]\Rightarrow2=17.6t-4.9t^2[/tex]subtracting 2 from both sides gives
[tex]17.6t-4.9t^2-2=0[/tex]The above can be rewritten in a more familiar form by a bit of rearranging of the terms.
[tex]-4.9t^2+17.6t-2=0[/tex]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:
[tex]t=\frac{-17.6\pm\sqrt[]{17.6^2-4(-4.9)(-2)}}{2\times(-4.9)}[/tex][tex]\begin{gathered} \Rightarrow t=0.1175 \\ t=3.474 \end{gathered}[/tex]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.
