Projectile motion equation stuck downwards velocity and angle

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A projectile is fired horizontally at 150 m/s from a 196 m high cliff, resulting in a time of flight of 6.32 seconds and a horizontal range of 948 m. The vertical component of the velocity upon impact is calculated to be approximately 62 m/s downward. The velocity vector is expressed as 150i - 62j, indicating both horizontal and vertical components. The magnitude of the velocity can be derived from this vector, which is essential for understanding the projectile's impact speed. The discussion highlights the importance of distinguishing between velocity and speed in projectile motion calculations.
emvizzle
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1. A projectile motion is fired horizontally at 150ms-1 from the top of a 196m high cliff. Calculate its velocity on hitting the ground.



2. Okay so I've already discovered its time of flight t = 6.32s and its range Δx = 948m.
Other figures are Δy= 196m, ay= -9.8ms-1, ux=150ms-1, Vy= 0ms-1, uy= 61.96ms-1




 
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Hey.All your answers are spot on .

The velocity vector is 150i-62j

(rounding 61.96 as 62).

So you have the velocity vector.You can find the magnitude of velocity (which I think the question meant by asking the velocity.Speed would have been a better term)and the angle it makes with the horizontal.
 
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