A person who is 1.5 m tall throws a ball horizontally with a speed of 7.8 m/s...
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A person who is 1.5 m tall throws a ball horizontally with a speed of 7.8 m/s...

[From: ] [author: ] [Date: 12-07-12] [Hit: ]
In a vertical direction, initial vertical velocity component, u =0.v² = 0² +2(-9.8)(-1.v = √29.......
What is the speed of the ball the moment before it strikes the ground (in m/s)?

(projectile motion problem)

I don't know the angle... so I don't know what to do :(

-
You don't need the angle.

Find the vertical speed component (v) on hitting the ground.
v² = u² +2as (or vf² = vi² + 2as if you use those symbols)
In a vertical direction, initial vertical velocity component, u =0. Taking up as positive:
v² = 0² +2(-9.8)(-1.5)
v = √29.4 = 5.42m/s

To get the total speed, add the vertical and horizontal components using vector addition (Pythagoras):
V = √(5.42² + 7.8²) = 9.5m/s

There 's a very similar calculation in the link at about 8mins 18 seconds.
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