Using kinematic equation, v^2 - u^2 = 2as. 5^2 - 3^2 = 2a x 16. a = 0.5m/s^2. So particle will deaccelerate at 0.5m/s^2. ( v = final velocity, u= initial velocity, a= acceleration, s= displacement.)
Answer:
2500/240 will give the resistance to be 10.42
Answer:
The velocity at the top of its path will be zero (0)
Explanation:
We can solve this problem or particular situation using the principle of energy conservation.
Which tells us that energy is transformed from kinetic energy to potential energy and vice versa. A reference point should be considered at which the potential energy is zero, and at this point the initial velocity of 40 [m/s] is printed to the ball.
![Ek=Ep\\where:\\Ek=kinetic energy [J]\\Ep=potencial energy [J]](https://tex.z-dn.net/?f=Ek%3DEp%5C%5Cwhere%3A%5C%5CEk%3Dkinetic%20energy%20%5BJ%5D%5C%5CEp%3Dpotencial%20energy%20%5BJ%5D)
The potential energy is determined by:
![Ep=m*g*h\\where:\\m=mass of the ball[kg}\\g=gravity[m/s^2]\\h=heigth [m]\\](https://tex.z-dn.net/?f=Ep%3Dm%2Ag%2Ah%5C%5Cwhere%3A%5C%5Cm%3Dmass%20of%20the%20ball%5Bkg%7D%5C%5Cg%3Dgravity%5Bm%2Fs%5E2%5D%5C%5Ch%3Dheigth%20%5Bm%5D%5C%5C)
The kinetic energy is determined by:
![Ek=\frac{1}{2}*m*v_{0} ^{2} \\where\\v_{0} = initial velocity[m/s]](https://tex.z-dn.net/?f=Ek%3D%5Cfrac%7B1%7D%7B2%7D%2Am%2Av_%7B0%7D%20%5E%7B2%7D%20%20%5C%5Cwhere%5C%5Cv_%7B0%7D%20%3D%20initial%20velocity%5Bm%2Fs%5D)
![Ek=Ep\\\frac{1}{2} *m*v_{0} ^{2} =m*9.81*h\\h=\frac{40^{2}}{2*9.81} \\h=81.5[m]](https://tex.z-dn.net/?f=Ek%3DEp%5C%5C%5Cfrac%7B1%7D%7B2%7D%20%2Am%2Av_%7B0%7D%20%5E%7B2%7D%20%3Dm%2A9.81%2Ah%5C%5Ch%3D%5Cfrac%7B40%5E%7B2%7D%7D%7B2%2A9.81%7D%20%5C%5Ch%3D81.5%5Bm%5D)
This will be the maximum path but, its velocity at this point will be zero. Because now all the kinetic energy has been transformed in potential energy.