Question

A projectile is launched at an angle above the
ground. The horizontal component of the
projectile’s velocity, vx, is initially 40. meters per
second. The vertical component of the projectile’s
velocity, vy, is initially 30. meters per second.
What are the components of the projectile’s
velocity after 2.0 seconds of flight? [Neglect
friction.]
(1) vx = 40. m/s and v
y = 10. m/s
(2) vx = 40. m/s and v
y = 30. m/s
(3) vx = 20. m/s and v
y = 10. m/s
(4) vx = 20. m/s and v
y = 30. m/s

Answer 1

The first rule of vectors is that the horizontal and vertical components are separate. Disregarding air resistance, the only thing we have to worry about is gravity.

The appropriate suvat to use for the vertical component is v = u +at
I will take a to be -9.81, you may have to change it to be 10 if your qualification likes g to be 10.

v = 30 + (-9.81x2)
v = 30 - 19.62
=10.38m/s

Therefore we know that after 2.0 s the vertical component will be 10.38ms^-1, ie 10m/s as the answers given are all to 2sf.

The horizontal component is completely separate to the vertical component and since there is no air resistance, it will remain constant throughout the projectiles trajectory. Therefore it will remain at 40ms^-1.

Combining this together we get:
(1) vx=40m/s and vy=10m/s

Answer 2

The horizontal component of the velocity after 2 seconds is 40 m/s and the vertical component of the velocity is 10.4 m/s.

The given parameters;

• initial horizontal velocity of the projectile, $v_x$ = 40 m/s
• initial vertical velocity of the projectile, $v_y$ = 30 m/s
• time of motion, t = 2s

The vertical component of the velocity after 2 seconds is calculated as follows;

$v_f_y = v_o _y - gt\\\\v_f_y = 30 - (2\times 9.8)\\\\v_f_y = 10.4 \ m/s$

The horizontal component of the velocity after 2 seconds is calculated as follows;

$v_f_x = v_0_x \\\\v_f_x = 40 \ m/s$

The horizontal velocity is not affected by gravity.

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