Answer to Question #135060 in Physics for vala

Question

A stone is thrown horizontally with an initial speed of 30 m/s from a bridge. Find the stone’s total speed when it enters the water 4 seconds later. (Ignore air resistance.)

(A) 30 m/s (B) 40 m/s (C) 50 m/s (D) 60 m/s (E) 70 m/s 

Answer 1

Let's choose the upwards as the positive direction and find vertical ($y$-component) of the stone's speed from the kinematic equation:

$$v_y = v_{0y} +gt,$$

here, $v_{0y} = 0 \ ms^{-1}$ is the initial vertical speed of the stone, $g = -10 ms^{-2}$ is the acceleration due to gravity, $t = 4 \ s$ is the time during which the stone flies from the bridge to the water.

Then, we get:

$$v_y = -10 \ ms^{-2} \cdot 4 \ s = -40 \ ms^{-1}.$$

The sign minus indicates that the vertical component of the stone's speed directed downward.

Finally, we can find the stone’s total speed from the Pythagorean theorem:

$$v = \sqrt{v_x^2+v_y^2},$$

here, $v_x = 30 \ ms^{-1}$ is the horizontal component of the stone's speed.

Then, we get:

$$v = \sqrt{(30 \ ms^{-1})^2+(-40 \ ms^{-1})^2} = 50 \ ms^{-1}.$$

Answer:

(C) $v = 50 \ ms^{-1}.$