Answer:
Time = 1.75[s]; Distance traveled = 21.5 [m]; Max height = 15 [m]
Explanation:
First, we have to break down the velocity vector into the X & y components.
![(v_{x})_{0} = 15 * cos( 35)= 12.28[m/s]\\(v_{y})_{0} = 15 * sin( 35)= 8.6[m/s]\\\\](https://tex.z-dn.net/?f=%28v_%7Bx%7D%29_%7B0%7D%20%3D%2015%20%2A%20cos%28%2035%29%3D%2012.28%5Bm%2Fs%5D%5C%5C%28v_%7By%7D%29_%7B0%7D%20%3D%2015%20%2A%20sin%28%2035%29%3D%208.6%5Bm%2Fs%5D%5C%5C%5C%5C)
To find the time t that lasts the ball of cannon in the air we must use the following equation of kinematics, in this equation the value of y is equal to zero because it will be proposed that the ball lands at the same level that was fired.
![y=(v_{y} )_{0}-\frac{1}{2}*g*t^{2} \\where:\\g=9.81[m/s^2]\\t = time[s]\\y=0[m]](https://tex.z-dn.net/?f=y%3D%28v_%7By%7D%20%29_%7B0%7D-%5Cfrac%7B1%7D%7B2%7D%2Ag%2At%5E%7B2%7D%20%20%20%5C%5Cwhere%3A%5C%5Cg%3D9.81%5Bm%2Fs%5E2%5D%5C%5Ct%20%3D%20time%5Bs%5D%5C%5Cy%3D0%5Bm%5D)
![0=8.6*t-\frac{1}{2}*9.81*t^{2} \\4.905*t^{2}=8.6*t\\ t=1.75[s]](https://tex.z-dn.net/?f=0%3D8.6%2At-%5Cfrac%7B1%7D%7B2%7D%2A9.81%2At%5E%7B2%7D%20%20%5C%5C4.905%2At%5E%7B2%7D%3D8.6%2At%5C%5C%20t%3D1.75%5Bs%5D)
In order to find the distance traveled horizontally from the cannonball, we must use the speed kinematics equation in the X coordinate.
![x = (v_{x})_{0} *t\\x=12.28*1.75\\x=21.5 [m]](https://tex.z-dn.net/?f=x%20%3D%20%28v_%7Bx%7D%29_%7B0%7D%20%20%2At%5C%5Cx%3D12.28%2A1.75%5C%5Cx%3D21.5%20%5Bm%5D)
In order to find the last value, we must bear in mind that when the cannonball reaches the maximum height, the velocity in the component y is equal to zero, and we can find the value of and with the following kinematic equation
![y = (v_{y})_{0} *t+\frac{1}{2} *g*(t)^{2} \\y = 0*t+\frac{1}{2} *9.81*(1.75)^{2}\\ y=15 [m]](https://tex.z-dn.net/?f=y%20%3D%20%28v_%7By%7D%29_%7B0%7D%20%2At%2B%5Cfrac%7B1%7D%7B2%7D%20%2Ag%2A%28t%29%5E%7B2%7D%20%5C%5Cy%20%3D%200%2At%2B%5Cfrac%7B1%7D%7B2%7D%20%2A9.81%2A%281.75%29%5E%7B2%7D%5C%5C%20y%3D15%20%5Bm%5D)
Answer:
2872.8 N
Explanation:
We have the following information
m =n72kg
Δy = 18m
t = 0.95s.
From here we use the equation
Δy=1/2at2 in order to solve for the acceleration.
So a
=( 2x 18m)/(0.95s²)
= 36/0.9025
= 39.9m/s2.
From there we use the equation
F = ma
F=(72kg) x (39.9)
= 2872.8N.
2872.8N is the average net force exerted on him in the barrel of the cannon.
Thank you!
Answer:
Explanation:
<u>Instant Velocity and Acceleration
</u>
Give the position of an object as a function of time y(x), the instant velocity can be obtained by
Where y'(x) is the first derivative of y respect to time x. The instant acceleration is given by
We are given the function for y
Note we have changed the last term to be quadratic, so the question has more sense.
The velocity is
And the acceleration is
The object will move 5 meters per second (5m/s)
The chemical energy of petrol is converted to heat energy on combustion. The heat energy is converted to kinetic energy by the use of internal combustion engines in vehicles. The law of conservation of energy is maintained in each process.
<h3>What is kinetic energy?</h3>
The kinetic energy of an object is associated with its motion. It can be related to the mass and velocity as
K.E = 1/2 mv²
Given is a diagram of energy conversion due to combustion.
The chemical energy of petrol is converted to heat energy on combustion. The heat energy is converted to kinetic energy by the use of internal combustion engines in vehicles.
The law of conservation of energy states that the energy can neither be created nor destroyed. It can be only converted to one form to the other.
In the given process, the mass decreases but energy remains the same in all forms of energy.
Hence, the law of conservation of energy is maintained in each process
Learn more about kinetic energy.
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