Mechanics Projectile Questions

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A particle is projected horizontally from point A at (5i + 30j) m and reaches point B at (17i + 10.4j) m. The horizontal and vertical components of its motion are analyzed, leading to the calculation of the distance traveled using the Pythagorean theorem. The time taken to travel from A to B is determined to be 2 seconds. Using the formula v = s / t, the speed of projection is calculated to be 6 m/s. The solution confirms the correct application of kinematic equations for projectile motion.
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Homework Statement


At time t = 0 a particle is projected horizontally from a point A with position vector (5i + 30j)m. It passes through point B with position vector (17i +10.4j)m. Find:
- Its speed of projection
- Time taken to travel from A to B

Homework Equations



s = ut + 0.5at²

v² = u² + 2as

v = s / t

a = (v - u) / t

v = s / t

The Attempt at a Solution



Well from A to B the i and j components are

12i
-19.6j

i can draw a right angled triangle and use pythagerous's thyrom and calculate that the other side is 12.8 (3.s.f) - is that meters?

Have i gone the right way. I don't know what to do now. Any hints
 
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Ahh i worked out the verticle component and got t = 2

now...
 
then put this into v = s / t

v = 6m/s and t = 2s

yeh :)
 

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