Projectile motion equation stuck downwards velocity and angle

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SUMMARY

The discussion focuses on calculating the velocity of a projectile fired horizontally at 150 m/s from a height of 196 m. The time of flight is determined to be 6.32 seconds, with a horizontal range of 948 m. The velocity vector upon impact is calculated as 150i - 62j, leading to a magnitude of velocity and an angle with the horizontal that can be derived from these components. The distinction between velocity and speed is also noted, emphasizing the importance of terminology in physics problems.

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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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