Hi I need some physics help!?
The question is:

What force must a track star exert on a shot put of mass 5 kg to accelerate it from rest to a velocity of 6 m/s (up) while pushing it through a vertical distance of 80 cm? How long will it rise after leaving the shot-putter's hand?
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answers:
oldschool say: What force must a track star exert on a shot put of mass 5 kg to accelerate it from rest to a velocity of 6 m/s (up) while pushing it through a vertical distance of 80 cm? How long will it rise after leaving the shot-putter's hand?

We know F = m(a+g). If a = 0 the force is just the weight. If the mass is falling at g then a = -g and F = 0
Use Vf² = Vi² + 2*a*d -----> 6² = 0² + 2*a*0.8 -----> a = 36/1.6 = 22.5m/s²
F = 5(22.5+9.8) = 161.5 or about 162N <<<<<
Vy(t) = Vyo - g*t = 0 at t = 6/9.8 = 0.61s <<<<<
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oubaas say: (Vf^2-Vi^2)+2gh = 2*a*h
(6^2-0)+1.6*9.806 = a*1.6
acceler. a = (36+15.69)/1.6 = 32.3 m/sec^2
needed force F = m*a = 5*32.3 = 161.5 N

Δh = Vf^2/2g = 36/19.612 = 1.84 m
tup = V/g = 6/9.806 = 0.61 sec
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electron1 say: The first step is to use the following equation to determine the upward acceleration of the shot.

vf^2 = vi^2 + 2 * a * d, vi = 0 m/s, d = 0.8 m
6^2 = 2 * a * 0.8
a = 22.5 m/s^2

The net vertical force on the shot is the upward force exerted by the track star minus the weight of the shot.

Weight = 5 * 9.8 = 49 N

F – 49 = 5 * 22.5
F = 161.5 N

When it leaves the shot-putter's hand, it has an upward velocity of 6 m/s. At this point, the only force is its weight. This causes a downward acceleration of 9.8 m/s^2. Let’s use the following equation to determine the time for the shot to have a vertical velocity of 0 m/s.


vf = vi + g * t
0 = 6 + -9.8 * t
t = 6 ÷ 9.8

The time is approximately 0.61 second. I hope this is helpful for you.
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JosephV say: v² - u² = 2as
a = v²/2s = 36/(2*0.8) m/s²
Force = mass * acceleration = (5 * 36) / (1.6) Newton.
Please solve.
Part 2.
v² - u² = 2as
0 - 36 = 2*(-9.8)*s
s= 36/19.6 meters Please solve
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