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Maurinko [17]
2 years ago
12

6. An aircraft is is travelling along a runway at a Velocity of 25m/s in It accelerates at a rate of 4m/s² for a distance of 750

m before taking off.
Calculate its take-off speed
Physics
1 answer:
zysi [14]2 years ago
7 0

Answer:

Take-off velocity = v = 81.39[m/s]

Explanation:

We can calculate the takeoff speed easily, using the following kinematic equation.

v_{f}^{2}=v_{i}^{2} +2*a*x

where:

a = acceleration = 4[m/s^2]

x = distance = 750[m]

vi = initial velocity = 25 [m/s]

vf = final velocity

v_{f}=\sqrt{(25)^{2}+(2*4*750) } \\v_{f}=81.39[m/s]

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from the question you can see that some detail is missing, using search engines i was able to get a similar question on "https://www.slader.com/discussion/question/a-student-throws-a-water-balloon-vertically-downward-from-the-top-of-a-building-the-balloon-leaves-t/"

here is the question : A student throws a water balloon vertically downward from the top of a building. The balloon leaves the thrower's hand with a speed of 60.0m/s. Air resistance may be ignored,so the water balloon is in free fall after it leaves the throwers hand. a) What is its speed after falling for 2.00s? b) How far does it fall in 2.00s? c) What is the magnitude of its velocity after falling 10.0m?

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(A) 26 m/s

(B) 32.4 m

(C) v = 15.4 m/s

Explanation:

initial speed (u) = 6.4 m/s

acceleration due to gravity (a) = 9.9 m/s^[2}

time (t) = 2 s

(A)   What is its speed after falling for 2.00s?

  from the equation of motion v = u + at we can get the speed

v = 6.4 + (9.8 x 2) = 26 m/s

(B) How far does it fall in 2.00s?

  from the equation of motion s=ut+0.5at^{2} we can get the distance covered

s = (6.4 x 2) + (0.5 x 9.8 x 2 x 2)

s = 12.8 + 19.6 = 32.4 m

c) What is the magnitude of its velocity after falling 10.0m?

from the equation of motion below we can get the velocity

v^{2} = u^{2} + 2as\\v^{2} = 6.4^{2} + (2x9.8x10)\\V^{2} = 236.96\\v = \sqrt{236.96}

v = 15.4 m/s

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