Question

1. a boy standing on the bridge kicks a stone into the water below. he kicks the stone with a horizontal velocity of 3.28 m/s. it lands in the water 27.2 m away from the bridge. how high is the bridge?

Explanation:

Step1: Find time of flight

In horizontal motion (constant velocity), $v_x = \frac{x}{t}$, so $t = \frac{x}{v_x}$.
Given $v_x = 3.28\ m/s$, $x = 27.2\ m$.
$t = \frac{27.2}{3.28} \approx 8.29\ s$.

Step2: Find bridge height (vertical motion, free fall)

Vertical displacement $y = \frac{1}{2}gt^2$ (initial vertical velocity $u_y = 0$), $g = 9.8\ m/s^2$.
$y = \frac{1}{2} \times 9.8 \times (8.29)^2$.
$y = 4.9 \times 68.7241 \approx 336.75\ m$.

Answer:

The height of the bridge is approximately $\boldsymbol{337\ m}$ (or $336.75\ m$ depending on precision).