Question

A small plane takes off at a constant velocity of 175 km/h at an angle of 35.5°. At 48.0 s,Question 14What is the velocity of the plane in m/s?Round your answer to 3 significant figures.Question 15How far has its shadow traveled from liftoff along the ground ? Assume the sun is directly above the plane.Add you

Answer 1

Given data

(14)

*The given constant velocity of the plane take off is v = 175 km/ h = 48.61 m/s

*The given angle is

`\theta=35.5^0`

*The given time is t = 48.0 s

The vertical component of the velocity is calculated as

`\begin{gathered} v_y=v_{}\sin \theta-gt \\ =(48.61)*\sin 35.5-(9.8)(48.0) \\ =-442.17\text{ m/s} \end{gathered}`

The horizontal component of the velocity is calculated as

`\begin{gathered} v_x=v\cos \theta \\ =(48.61)*\cos 35.5^0 \\ =39.57\text{ m/s} \end{gathered}`

The formula for the velocity of the plane is given as

`v=\sqrt[]{(v_x_{})^2+(v_y)^2_{}}`

Substitute the known values in the above expression as

`\begin{gathered} v=\sqrt[]{(39.57)^2+(-442.17)^2} \\ =443.93\text{ m/s} \end{gathered}`

Hence, the velocity of the plane in meter per second (m/s) is v = 443.93 m/s