Question

question 43
(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 magnitude of the vertical displacement of the projectile?
150 m
15 m
29 m
490 m
44 m

Explanation:

Step1: Identify the vertical motion formula

For vertical motion (initial vertical velocity \(v_{0y} = 0\) as fired horizontally), use \(y=v_{0y}t+\frac{1}{2}gt^{2}\). Here \(g = 9.8\ m/s^{2}\), \(v_{0y}=0\ m/s\), \(t = 3.00\ s\).

Step2: Substitute values into the formula

Substitute \(v_{0y}=0\), \(t = 3\ s\), \(g=9.8\ m/s^{2}\) into \(y=v_{0y}t+\frac{1}{2}gt^{2}\). We get \(y=0\times3+\frac{1}{2}\times9.8\times3^{2}\).
Calculate \(\frac{1}{2}\times9.8\times9=4.9\times9 = 44.1\approx44\ m\)

Answer:

44 m