A projectile is fired from point 0 at the edge of a cliff, with initial velocity

Question

A projectile is fired from point 0 at the edge of a cliff, with initial velocity components of V0x = 60 m/s and V0y = 175 m/s. The projectile rises and then falls into the sea at point P. The time of flight of the projectile is 40.0 s, and it experiences no appreciable air resistance in flight. What is the magnitude of the velocity of the projectile 21.0 s after it is fired?

From everything that I am looking at it seems that V0x = 60i but then V0y is calculated as V0y = 175-9.8(21) = -30.8j, and then sqrt((60)^2+(-30.8)^2). So if this is correct why are these calculations done for V0y but not for V0x?

Explanation / Answer

first

vx = 60 m/s

vy = v0y + a*t

vy = 175 - 9.8 * 21 = -30.8 m/s

v = sqrt( 60^2 + (-30.8)^2)

v = 67.4 m/s

so the velocity after 21 sec is 67.4 m/s

vx is the horizontal velocity and vy is the vertical velocity

so in the vertical motion there is gravitational force acting on it so that's why we use the calculation for vy

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