Question

question 45
(problem reference m.5)
a projectile is fired horizontally with an initial speed of 50.0 m/s. it strikes the ground 3.00 s later. neglect air resistance.
what is the vertical component of the velocity of the projectile with which its strikes the ground?
50 m/s
3.3 m/s
79 m/s
29 m/s
zero m/s

Explanation:

Step1: Identify the formula

Use the kinematic equation \(v = v_0+at\). For vertical motion, the initial vertical velocity \(v_{0y}=0\) (fired horizontally), \(a = g= 9.8\ m/s^{2}\), and \(t = 3.00\ s\).

Step2: Calculate the vertical component of velocity

Substitute the values into the formula \(v_y=v_{0y}+gt\). Since \(v_{0y} = 0\), we have \(v_y=0+(9.8\ m/s^{2})\times3.00\ s\).

$$v_y=29.4\ m/s\approx29\ m/s$$

Answer:

29 m/s