Question

A ball is thrown straight up. It passes a b)00-m-high window 7.50 m off the ground on its path up and takes 0.312 s to go past the window. What was the ball’s initial velocity?

Calculate the ball's velocity just before it reaches the bottom of the window.
a) v_{{bottom}} = sqrt{2gh_{{bottom}}}
b) v_{{bottom}} = frac{2gh_{{bottom}}}{t_{{window}}}
c) v_{{bottom}} = frac{g cdot t_{{window}}}{2h_{{bottom}}}
d) v_{{bottom}} = frac{2h_{{bottom}}}{gt_{{window}}}

Answer 1

To find the initial velocity of the ball, first calculate the velocity just before it reaches the bottom of the window using the equation v_bottom = sqrt(2gh_bottom). Then, use the velocity at the bottom of the window to calculate the initial velocity using the equation v_bottom = (g * t_window) / 2h_bottom.

Step-by-step explanation:

To calculate the ball's initial velocity, we can first calculate the velocity just before it reaches the bottom of the window. We can use the equation: vbottom = √(2ghbottom) where g is the acceleration due to gravity (9.8 m/s²) and hbottom is the height of the window above the ground (2.00 m). Plugging in the values: vbottom = √(2 * 9.8 * 2.00) = 6.26 m/s.

Now, we can calculate the initial velocity of the ball using the equation: vbottom = (g * twindow) / 2hbottom where twindow is the time it takes for the ball to pass the window (0.312 s). Plugging in the values: 6.26 = (9.8 * 0.312) / (2 * 7.50) = 0.1633 m/s. Therefore, the ball's initial velocity is approximately 0.1633 m/s.