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question 12
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a projectile is fired horizontally with an initial speed of 57 m/s. what are the horizontal and vertical components of its displacement 3.0 s after it is fired?
o x = 170 m y = - 44 m
o x = 44 m y = -29 m
o x = 210 m y = zero m
o x = 210 m y = - 44 m
o x = 170 m y = zero m

Explanation:

Step1: Calculate horizontal displacement

The horizontal - motion of a projectile is a uniform - motion with constant velocity. The formula for horizontal displacement $x$ is $x = v_{0x}t$, where $v_{0x}$ is the initial horizontal velocity and $t$ is the time. Given $v_{0x}=57\ m/s$ and $t = 3.0\ s$, then $x=v_{0x}t=57\times3 = 171\approx170\ m$.

Step2: Calculate vertical displacement

The vertical - motion of a projectile is a free - fall motion with an initial vertical velocity $v_{0y}=0\ m/s$. The formula for vertical displacement $y$ is $y=v_{0y}t-\frac{1}{2}gt^{2}$, where $g = 9.8\ m/s^{2}$ and $t = 3.0\ s$. Since $v_{0y}=0\ m/s$, then $y=-\frac{1}{2}gt^{2}=-\frac{1}{2}\times9.8\times3^{2}=-44.1\approx - 44\ m$.

Answer:

$x = 170\ m\ \ y=-44\ m$ (First option)