The magnitude of the unknown height of the projectile is determined as 16.1 m.
<h3>
Magnitude of the height</h3>
The magnitude of the height of the projectile is calculated as follows;
H = u²sin²θ/2g
H = (36.6² x (sin 29)²)/(2 x 9.8)
H = 16.1 m
Thus, the magnitude of the unknown height of the projectile is determined as 16.1 m.
Learn more about height here: brainly.com/question/1739912
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The resultant vector is 11√2 km due north east.
<h3><u>Explanation:</u></h3>
The vector is a type of quantity which has both magnitude and direction. This quantities when expressed needs to specify both magnitude and direction.
We need to calculate the magnitude and direction separately.
Here firstly for the magnitude,
The magnitudes are both 11 km and they are at right angles to each other.
So, the resultant magnitude = √(11² +11²) km
=11√2 km
Now for the direction, one vector is due north and the other is due east.
So the resultant vector is due north east.
So the final vector is 11√2 km due North-East.
Answer:
v = 5.34[m/s]
Explanation:
In order to solve this problem, we must use the theorem of work and energy conservation. This theorem tells us that the sum of the mechanical energy in the initial state plus the work on or performed by a body must be equal to the mechanical energy in the final state.
Mechanical energy is defined as the sum of energies, kinetic, potential, and elastic.
E₁ = mechanical energy at initial state [J]

In the initial state, we only have kinetic energy, potential energy is not had since the reference point is taken below 1.5[m], and the reference point is taken as potential energy equal to zero.
In the final state, you have kinetic energy and potential since the car has climbed 1.5[m] of the hill. Elastic energy is not available since there are no springs.
E₂ = mechanical energy at final state [J]

Now we can use the first statement to get the first equation:

where:
W₁₋₂ = work from the state 1 to 2.


where:
h = elevation = 1.5 [m]
g = gravity acceleration = 9.81 [m/s²]

![58 = v^{2} +29.43\\v^{2} =28.57\\v=\sqrt{28.57}\\v=5.34[m/s]](https://tex.z-dn.net/?f=58%20%3D%20v%5E%7B2%7D%20%2B29.43%5C%5Cv%5E%7B2%7D%20%3D28.57%5C%5Cv%3D%5Csqrt%7B28.57%7D%5C%5Cv%3D5.34%5Bm%2Fs%5D)
Answer:
Below
Explanation:
You can use this equation to find the distance :
distance = velocity x time
distance = (26.7)(3.06)
= 81.702 m
Rounding to 3 sig figs
= 81.7 m
Hope this helps