The magnitude of the unknown height of the projectile is determined as 16.1 m.
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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.
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Answer: The average speed is 27,24 mph (exactly 1008/37 mph)
Explanation:
This is solved using a three rule: We know the speeds and the distances, what we can obtain from it is the time used. It is done like this:
1h--->18mi
X ---->20 mi, then X=20mi*1h/18mi= 10/9 h=1,111 h
1h--->56mi
X ---->20 mi, then X=20mi*1h/56mi= 5/14 h=0,35714 h
Then the average speed is calculated by taking into account that it was traveled 40mi and the time used was 185/126 h=1,468 h and since speed is distance over time we get the answer. Average speed= 40mi/(185/126 h)=1008/37 mph=27,24 mph.
Answer:
v = 29.4 m / s
Explanation:
For this exercise we can use the conservation of mechanical energy
Lowest starting point.
Em₀ = K = ½ m v²
final point. Higher
= U = m g h
Let's use trigonometry to lock her up
cos 60 = y / L
y = L cos 60
Height is the initial length minus the length at the maximum angle
h = L - L cos 60
h = L (1- cos 60)
energy is conserved
Em₀ = Em_{f}
½ m v² = mgL (1 - cos 60)
v = 2g L (1- cos 60)
let's calculate
v² = 2 9.8 3.0 (1- cos 60)
v = 29.4 m / s