Questions: To celebrate a victory, a pitcher throws her glove straight upward with an initial speed of 6.5 m / s. (a) How much time does it take for the glove to return to the pitcher? (b) How much time does it take for the glove to reach its maximum height?

Solution Steps

We need to determine the time it takes for a glove thrown upward to return to the pitcher and the time it takes to reach its maximum height. The initial speed of the glove is given as \(6.5 \, \text{m/s}\).

The motion of the glove can be analyzed using the equations of uniformly accelerated motion. The acceleration due to gravity (\(g\)) is \(9.8 \, \text{m/s}^2\).

At maximum height, the final velocity (\(v\)) of the glove is \(0 \, \text{m/s}\). Using the equation: \[ v = u + at \] where \(u\) is the initial velocity, \(a\) is the acceleration, and \(t\) is the time, we can solve for \(t\): \[ 0 = 6.5 \, \text{m/s} - (9.8 \, \text{m/s}^2) \cdot t \] \[ t = \frac{6.5 \, \text{m/s}}{9.8 \, \text{m/s}^2} \] \[ t = 0.6633 \, \text{s} \]

The total time for the glove to return to the pitcher is twice the time it takes to reach the maximum height, as the time to ascend and descend are equal: \[ t_{\text{total}} = 2 \times 0.6633 \, \text{s} \] \[ t_{\text{total}} = 1.3266 \, \text{s} \]

Final Answer