Explanation:
I believe <u>f</u><u>i</u><u>l</u><u>m</u><u> </u><u>i</u><u>s</u><u> </u>not part of a circuit
Answer:
The height of the building is 88.63 m.
Explanation:
Given;
initial component of vertical velocity, 
= 12 m/s sin 26° = 5.26 m/s
initial horizontal component of the velocity, 
= 12 m/s cos 26° =10.786 m/s
horizontal distance traveled by the rock, x = 40.4 m
time of flight is calculated as;
x = t
t = x /
t = 40.4 / 10.786
t = 3.75 s
Determine the final vertical velocity of the ball;
Determine the height of the rock;
Therefore, the height of the building is 88.63 m.
Volume = 873 - 50 = 723 cm^3
Answer:
V₀y = 0 m/s
t = 2.47 s
V₀ₓ = 61.86 m/s
Vₓ = 61.86 m/s
Explanation:
Since, the ball is hit horizontally, there is no vertical component of velocity at initial point. So, the initial vertical velocity (V₀y) will beL
<u>V₀y = 0 m/s</u>
For the initial vertical velocity of golf ball we consider the vertical motion and apply 2nd equation of motion:
Y = V₀y*t + (0.5)gt²
where,
Y = Height = 30 m
g = 9.8 m/s²
t = time to hit the ground = ?
Therefore,
30 m = (0 m/s)(t) + (0.5)(9.8 m/s²)t²
t² = 30 m/4.9 m/s²
t = √6.122 s²
<u>t = 2.47 s</u>
For initial vertical velocity we analyze the horizontal motion of the ball. We neglect the frictional effects in horizontal motion thus the speed remains uniform. Hence,
V₀ₓ = Xt
where,
V₀ₓ = Initial vertical Velocity = ?
X = Horizontal Distance = 25 m
Therefore,
V₀ₓ = (25 m)(2.47 s)
<u>V₀ₓ = 61.86 m/s</u>
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Due, to uniform motion in horizontal direction:
Final Vertical Velocity = Vₓ = V₀ₓ
Vₓ = 61.86 m/s
