Question

(problem reference m.4)
a projectile fired from a gun has initial horizontal and vertical components of velocity equal to 40 m/s and 60 m/s, respectively.
what is the magnitude of the projectiles velocity just before it strikes the ground. the ground it horizontal and level?
60 m/s
40 m/s
zero m/s
9.8 m/s
72 m/s

Explanation:

Step1: Analyze horizontal velocity

In projectile motion, the horizontal component of velocity \(v_x\) remains constant (assuming no air - resistance). Given \(v_{x0}=40\ m/s\), so \(v_x = 40\ m/s\) just before hitting the ground.

Step2: Analyze vertical velocity

The vertical motion is a free - fall motion. Using the kinematic equation \(v_y^2=v_{y0}^2 - 2g\Delta y\). Since the projectile starts and ends at the same height (\(\Delta y = 0\)), \(v_y=-v_{y0}\) (the negative sign indicates direction, but for magnitude \(|v_y| = |v_{y0}|\)). Given \(v_{y0}=60\ m/s\), so \(|v_y| = 60\ m/s\) just before hitting the ground.

Step3: Calculate the magnitude of the resultant velocity

The magnitude of the resultant velocity \(v\) is given by the Pythagorean theorem \(v=\sqrt{v_x^2 + v_y^2}\). Substitute \(v_x = 40\ m/s\) and \(v_y = 60\ m/s\) into the formula: \(v=\sqrt{40^{2}+60^{2}}=\sqrt{1600 + 3600}=\sqrt{5200}\approx72\ m/s\)

Answer:

72 m/s