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amm1812
3 years ago
11

The sun is 30 degrees above the horizon. it makes a 51 m long shadow of a tall tree. part a how high is the tree?

Physics
2 answers:
natali 33 [55]3 years ago
7 0
<span>Answer: Height of the tree = 29.44m

Explanation:
The Sun is 30 degrees above the horizon, and the shadow of a tree it makes is 51 meters long. The right angled triangle can be constructed, where the base of the triangle is actually the shadow of the tree (which is 51m long) and the perpendicular of a triangle is the height of the tree. By using trignometric equation:

tan(</span>Ф<span>) = perpendicular/base
tan(30) = height-of-a-tree/51
height-of-a-tree = tan(30) * 51
height-of-a-tree = 29.44m</span>
Lemur [1.5K]3 years ago
5 0

Answer:

Height of the tree must be 29.44 m

Explanation:

Here angle made by the horizon is

\theta = 30 degree

also we know that

tan\theta = \frac{H}{51}

now from above equation we have

tan30 = \frac{H}{51}

H = 51 tan30

now we have

H = 29.44 m

so height must be 29.44 m

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A man stands on the roof of a 15.0-m-tall building and throws a rock with a speed of 30.0 m&gt;s at an angle of 33.0%1b above th
Brilliant_brown [7]
The motion described here is a projectile motion which is characterized by an arc-shaped direction of motion. There are already derived equations for this type of motions as listed:

Hmax = v₀²sin²θ/2g
t = 2v₀sinθ/g
y = xtanθ + gx²/(2v₀²cos²θ)

where

Hmax = max. height reached by the object in a projectile motion
θ=angle of inclination
v₀= initial velocity
t = time of flight
x = horizontal range
y = vertical height

Part A. 

Hmax = v₀²sin²θ/2g = (30²)(sin 33°)²/2(9.81)
Hmax = 13.61 m

Part B. In this part, we solve the velocity when it almost reaches the ground. Approximately, this is equal to y = 28.61 m and x = 31.91 m. In projectile motion, it is important to note that there are two component vectors of motion: the vertical and horizontal components. In the horizontal component, the motion is in constant speed or zero acceleration. On the other hand, the vertical component is acting under constant acceleration. So, we use the two equations of rectilinear motion:

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a = (v₁-v₀)/t
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v₁ = 38.23 m/s

Part C. 
y = xtanθ + gx²/(2v₀²cos²θ)
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13.61 + 15 = xtan33° + (9.81)x²/[2(30)²(cos33°)²]
Solving using a scientific calculator,
x = 31.91 m

3 0
3 years ago
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