Answer:
Velocity=14[m/s]
Explanation:
We can solve this problem by using the principle of energy conservation, where potential energy becomes kinetic energy.
In the attached image we can see the illustration of the ball falling from the height of 20 meters, at this time the potential energy will have the following value.
![Ep=m*g*h\\where:\\m=3[kg]\\h=20[m]\\](https://tex.z-dn.net/?f=Ep%3Dm%2Ag%2Ah%5C%5Cwhere%3A%5C%5Cm%3D3%5Bkg%5D%5C%5Ch%3D20%5Bm%5D%5C%5C)
![Ep=3*9.81*20\\Ep=588.6[J]](https://tex.z-dn.net/?f=Ep%3D3%2A9.81%2A20%5C%5CEp%3D588.6%5BJ%5D)
When the ball passes through half of the distance (10m) its potential energy will have decreased by half as shown below.
![Ep=3*9.81*10\\Ep=294.3[m]](https://tex.z-dn.net/?f=Ep%3D3%2A9.81%2A10%5C%5CEp%3D294.3%5Bm%5D)
If we know that potential energy is transformed into kinetic energy, we can find the value of speed.
![Ek=\frac{1}{2} *m*v^{2} \\therefore\\v=\sqrt{\frac{Ek*2}{m} } \\v=\sqrt{\frac{294.3*2}{3} } \\\\v=14[m/s]](https://tex.z-dn.net/?f=Ek%3D%5Cfrac%7B1%7D%7B2%7D%20%2Am%2Av%5E%7B2%7D%20%5C%5Ctherefore%5C%5Cv%3D%5Csqrt%7B%5Cfrac%7BEk%2A2%7D%7Bm%7D%20%7D%20%5C%5Cv%3D%5Csqrt%7B%5Cfrac%7B294.3%2A2%7D%7B3%7D%20%7D%20%5C%5C%5C%5Cv%3D14%5Bm%2Fs%5D)
When lost fluid is not replaced adequately, dehydration can result.
Answer:
See explanation
Explanation:
a) maximum height of a projectile = u sin^2θ/2g
H= 600 × (sin 30)^2/2 × 10
H= 7.5 m
b) Time of flight
t= 2u sinθ/g
t= 2 × 600 sin 30/10
t= 60 seconds
Range
R= u^2sin2θ/g
R= (600)^2 × sin2(30)/10
R= 31.2 m
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Explanation:
