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BigorU [14]
3 years ago
14

an aircraft is flying at height of 3400m above the ground.if the angle subtended at a ground observation point by the aircraft p

ositions 10s apart is 30degree. what is the speed of the aircraft?
Physics
1 answer:
Kamila [148]3 years ago
7 0

Answer: 588.9 m/s

Explanation:

Given that :

θ = 30°

Height, h = 3400m

Time, t = 10 seconds

From trigonometry ;

Tanθ = opposite / hypotenus

Tan 30 = 3400 / x

x tan 30 = 3400

0.5773502x = 3400

x = 3400 / 0.5773502

x = 5888.9727

Recall ;

Speed = Distance / time

Speed = 5888.9727 / 10

Speed = 588.897 m/s

Speed = 588.9 m/s

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<h2>Answer: 277.777 m</h2>

Explanation:

The situation described here is parabolic movement. However, as we are told that the rock was<u> projected upward from the surface</u>, we will only use the equations related to the Y axis.

In this sense, the movement equations in the Y axis are:

y-y_{o}=V_{o}.t+\frac{1}{2}g.t^{2}    (1)

V=V_{o}-g.t    (2)

Where:

y  is the rock's final position

y_{o}=0  is the rock's initial position

V_{o}=30\frac{m}{s} is the rock's initial velocity

V is the final velocity

t is the time the parabolic movement lasts

g=1.62\frac{m}{s^{2}}  is the acceleration due to gravity at the surface of the moon

As we know y_{o}=0 , equation (2) is rewritten as:

y=V_{o}.t+\frac{1}{2}g.t^{2}    (3)

On the other hand, the maximum height  is accomplished when V=0:

V=V_{o}-g.t=0    (4)

V_{o}-g.t=0    

V_{o}=g.t    (5)

Finding t:

t=\frac{V_{o}}{g}    (6)

Substituting (6) in (3):

y=V_{o}(\frac{V_{o}}{g})+\frac{1}{2}g(\frac{V_{o}}{g})^{2}    (7)

y_{max}=\frac{{V_{o}}^{2}}{2g}    (8)  Now we can calculate the maximum height of the rock

y_{max}=\frac{{(30m/s)}^{2}}{(2)(1.62m/s^{2})}   (9)

Finally:

y_{max}=277.777m  

4 0
3 years ago
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