Question

a ball is launched from ground at an angle with an initial speed of 10 m/s ($v_x = 6$ m/s, $v_y = 8$ m/s). the ball travels over flat ground. ignore the effects of air resistance. what is the speed of the ball the instant before it hits the ground?
a 0.0 m/s
b 6 m/s
c 8 m/s
d 10 m/s

Explanation:

Step1: Analyze horizontal and vertical velocities

In projectile motion, the horizontal velocity \(v_x\) remains constant (since there is no acceleration in the horizontal direction when air - resistance is ignored). The vertical velocity \(v_y\) changes due to gravity. But when the ball hits the ground, the vertical velocity has the same magnitude as the initial vertical velocity (because of the symmetry of projectile motion in the absence of air - resistance, \(v_{y - final}=-v_{y - initial}\)), and the horizontal velocity \(v_x\) remains \(v_{x - initial}\).

The speed \(v\) of an object with horizontal velocity \(v_x\) and vertical velocity \(v_y\) is given by the Pythagorean theorem \(v = \sqrt{v_x^{2}+v_y^{2}}\)

We know that \(v_x = 6m/s\) (remains constant throughout the motion) and when the ball hits the ground, \(|v_y|=8m/s\) (same magnitude as the initial vertical velocity)

Step2: Calculate the speed

Substitute \(v_x = 6m/s\) and \(v_y = 8m/s\) into the formula \(v=\sqrt{v_x^{2}+v_y^{2}}\)

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Answer:

D. 10 m/s