Answer:
-26 m/s.
Explanation:
Hello,
In this case, since the vertical initial velocity is 26 m/s and the vertical final velocity is 0 m/s at P, we compute the time to reach P:

With which we compute the maximum height:

Therefore, the final velocity until the floor, assuming P as the starting point (Voy=0m/s), turns out:

Which is clearly negative since it the projectile is moving downwards the starting point.
Regards.
Answer:
-589.05 J
Explanation:
Using work-kinetic energy theorem, the work done by friction = kinetic energy change of the base runner
So, W = ΔK
W = 1/2m(v₁² - v₀²) where m = mass of base runner = 72.9 kg, v₀ = initial speed of base runner = 4.02 m/s and v₁ = final speed of base runner = 0 m/s(since he stops as he reaches home base)
So, substituting the values of the variables into the equation, we have
W = 1/2m(v₁² - v₀²)
W = 1/2 × 72.9 kg((0 m/s)² - (4.02 m/s)²)
W = 1/2 × 72.9 kg(0 m²/s² - 16.1604 m²/s²)
W = 1/2 × 72.9 kg(-16.1604 m²/s²)
W = 1/2 × (-1178.09316 kgm²/s²)
W = -589.04658 kgm²/s²
W = -589.047 J
W ≅ -589.05 J
Stop using it all the time for some useless things.
Answer:

3257806.62409 m/s
Explanation:
G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²
M = Mass of Sun = 
r = Radius of Star = 20 km
u = Initial velocity = 0
v = Final velocity
s = Displacement = 16 m
a = Acceleration
Gravitational acceleration is given by

The gravitational acceleration at the surface of such a star is 

The velocity of the object would be 3257806.62409 m/s
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