Answer:
96.7 s
Explanation:
Time of flight in projectile can be calculated thus:
T = 2 × u × sin ϴ/ g
Where;
T = time of flight (s)
u = initial velocity (m/s)
ϴ = Angle of projectile (°)
g = acceleration due to gravity (9.8m/s²)
Based on the provided information; u = 670m/s, ϴ = 45°
Hence, using T = 2.u.sin ϴ/ g
T = 2 × 670 × sin 45° ÷ 9.8
T = 1340 × 0.7071 ÷ 9.8
T = 947.52 ÷ 9.8
T = 96.68
T = 96.7s
Answer:
The correct option is propped cantilever beam.
Explanation:
A propped cantilever beam along with all the reactions is shown in the attached figure.
Since we can see that the number of unknowns are 4 but the equations of statics are only 3 thus we conclude that the beam is indeterminate to 1 degree.
For all the other beams the reactions and the equations are 3 each thus making the system determinate.
Answer:K.E=449598.5j
Explanation:
Kinetic energy of a moving car=1/2mv^2
Where m is the mass of the car
And V is the velocity of the car
K.E=1/2 ×1300×26.3^2
K.E=449598.5j
Answer:
See the answers below
Explanation:
We can solve this problem using the principle of energy conservation. That is, the energy is conserved before and after dropping the bag.
For this case we have mechanical energy, which is the sum of the kinetic and potential energies.

where:

Ek = kinetic energy [J] (units of Joules)
Ep = potential energy [J]
In the final Energy (2), there is only potential energy. since when the balloon reaches the maximum height its velocity is zero, that is, there is no kinetic energy.
A)
![m*g*h+\frac{1}{2}*m*v^{2} =m*g*h_{1} \\9.81*50+0.5*(15)^{2}=9.81*h_{1}\\h_{1} = 61.46 [m]](https://tex.z-dn.net/?f=m%2Ag%2Ah%2B%5Cfrac%7B1%7D%7B2%7D%2Am%2Av%5E%7B2%7D%20%3Dm%2Ag%2Ah_%7B1%7D%20%5C%5C9.81%2A50%2B0.5%2A%2815%29%5E%7B2%7D%3D9.81%2Ah_%7B1%7D%5C%5Ch_%7B1%7D%20%3D%2061.46%20%5Bm%5D)
B)
With the value calculated above we can find the acceleration of the balloon.
The distance traveled is the difference between the maximum height and 50 meters.
![x = 61.46-50\\x = 11.46[m]](https://tex.z-dn.net/?f=x%20%3D%2061.46-50%5C%5Cx%20%3D%2011.46%5Bm%5D)
With the following equation of kinematics.

![0 = 15^{2} +2*a*11.46\\a = - 9.816 [m/s^{2} ]](https://tex.z-dn.net/?f=0%20%3D%2015%5E%7B2%7D%20%2B2%2Aa%2A11.46%5C%5Ca%20%3D%20-%209.816%20%5Bm%2Fs%5E%7B2%7D%20%5D)
The negative sign indicates that the acceleration acts downward. That is, in the opposite direction to the movement.
We can use the following equation of kinematics to find the final velocity after 4 seconds.
![v_{f}=v_{o}-a*t\\v_{f}=15-9.816*(4)\\v_{f}=-24.24 [m/s]](https://tex.z-dn.net/?f=v_%7Bf%7D%3Dv_%7Bo%7D-a%2At%5C%5Cv_%7Bf%7D%3D15-9.816%2A%284%29%5C%5Cv_%7Bf%7D%3D-24.24%20%5Bm%2Fs%5D)
Now the distance:
![v_{f}^{2} =v_{o}^{2}-2*a*x\\(24.24)^{2} =(15)^{2} -2*9.81*x\\x = 18.48 [m]\\x_{f}=50+18.48 = 68.48 [m]](https://tex.z-dn.net/?f=v_%7Bf%7D%5E%7B2%7D%20%3Dv_%7Bo%7D%5E%7B2%7D-2%2Aa%2Ax%5C%5C%2824.24%29%5E%7B2%7D%20%3D%2815%29%5E%7B2%7D%20-2%2A9.81%2Ax%5C%5Cx%20%3D%2018.48%20%5Bm%5D%5C%5Cx_%7Bf%7D%3D50%2B18.48%20%3D%2068.48%20%5Bm%5D)
c) Using the following equation of kinematics.
![v_{f}=v_{o}-a*t\\0 = 15-9.81*t\\15=9.81*t\\t = 1.52 [s]](https://tex.z-dn.net/?f=v_%7Bf%7D%3Dv_%7Bo%7D-a%2At%5C%5C0%20%3D%2015-9.81%2At%5C%5C15%3D9.81%2At%5C%5Ct%20%3D%201.52%20%5Bs%5D)