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Answer:
The components of the object's velocity 3 seconds after launch are
[tex]v_x = 30[/tex] m/s and [tex]v_y=-9.43[/tex] m/s.
Explanation:
The given components of initial velocity are
[tex]v_x= 30[/tex] m/s,
[tex]v_y= 20[/tex] m/s.
Assuming the y-axis is along the direction of gravitational force, so, the velocity along the x-axis will remain unchanged.
So, the x-componebt of the velocity after 3 seconds, [tex]v_x = 30[/tex] m/s.
Using the equation of motion along the y-direction, the final velocity, v, of a mass after time, t, having initial velocity, u is
v=u+at
In this case, [tex]u=y_y= 20 m/s, a=-9.81 m/s^2[/tex] ( as upward direction is taken as positive)
So, at time, t=3s, the final velocity in y-direction is
[tex]v= 20+(-9.81)\times 3[/tex]
[tex]\Rightarrow v=20-29.43[/tex]
[tex]\Rightarrow v= -9.43 m/s[/tex]
So, the y-component of the final velocity is 9.43 m/s in the downward direction.
Hence, the components of the object's velocity 3 seconds after launch are
[tex]v_x = 30 m/s[/tex] and [tex]v_y=-9.43 m/s[/tex].
The final vertical velocity of the object after 3 seconds is 49.4 m/s.
The final horizontal velocity of the object is 30 m/s.
The given parameters;
- initial horizontal velocity, Vx = 30 m/s
- initial vertical velocity, Vy = 20 m/s
- time of motion, t = 3 s
The final vertical velocity of the object after 3 seconds is calculated as follows;
[tex]v_f_y = v_0_y + gt\\\\v_f_y = 20 + 3(9.8)\\\\v_f_y = 49.4 \ m/s[/tex]
The final horizontal velocity of the object is the same as the initial horizontal velocity because horizontal motion is not affected by gravity.
[tex]v_f_x = v_0_x = 30 \ m/s[/tex]
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