O celebrate a victory, a pitcher throws her glove straight upward with an initial speed of 6.0 m/s. (a) how long does it take the glove to return tothe pitcher? (b) how long does it take for the glove to reach its maximum height?

Respuesta :

As per the kinematics equation

[tex]\delta s = v_i * t + \frac{1}{2} at^2[/tex]

when glove comes back to his hand the displacement will become zero

[tex] 0 = v_i * t - \frac{1}{2} gt^2[/tex]

[tex]0 = 6 * t - 4.9 t^2[/tex]

[tex]t = 1.22 s[/tex]

Part b)

Now as per symmetric nature the time to reach the top will be same as the time to come back again

so here we can say

[tex]T = t + t[/tex]

[tex]1.22 = 2t[/tex]

[tex]t = 0.61 s[/tex]

so it took 0.61 s to reach the top