Question

question 22 4 points save answer (problem reference 3 - 3) a shell is fired with a horizontal velocity in the positive x - direction from the top of an 80 - m high cliff. the shell strikes the ground 1330 m from the base of the cliff. the drawing is not to scale. determine the y - component of the velocity of the shell at the instant when it hits the ground. 20 m/s 39 m/s 170 m/s 330 m/s

Explanation:

Step1: Identify vertical - motion equation

Use the equation $v_y^2 = v_{0y}^2+ 2a_y\Delta y$. The initial vertical velocity $v_{0y}=0$ m/s, acceleration $a_y = g= 9.8$ m/s², and $\Delta y=80$ m.

Step2: Substitute values

Since $v_{0y} = 0$, the equation simplifies to $v_y^2=2g\Delta y$. Substitute $g = 9.8$ m/s² and $\Delta y = 80$ m. So $v_y^2=2\times9.8\times80$.

Step3: Calculate $v_y$

$v_y^2=1568$. Then $v_y=\sqrt{1568}\approx39.6$ m/s $\approx39$ m/s.

Answer:

39 m/s