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The altitude of a hang glider is increasing at a rate of 6.79 m/s. At the same time, the shadow...

Question:

The altitude of a hang glider is increasing at a rate of 6.79 m/s. At the same time, the shadow of the glider moves along the ground at a speed of 14.8 m/s when the sun is directly overhead. Calculate the magnitude of the glider's velocity.

Velocity:

The kinematic quantity that represents the position traveled in a linear direction by the object in unit time is known as velocity. The measurable unit of velocity is meter per second.

Answer and Explanation:


Given Data:

  • The rate at which the velocity of hang glider increases is: {eq}{v_g} = 6.79\;{\rm{m/s}} {/eq}
  • The speed of shadow of the glider move along ground when the sun is directly overhead is: {eq}{v_s} = 14.8\;{\rm{m/s}} {/eq}


The speed of hang glider acts in vertical direction and speed of shadow of gilder along ground when the sun is directly overhead acts in horizontal direction.

The expression for magnitude of the glider's velocity is

{eq}{v_o} = \sqrt {v_g^2 + v_s^2} {/eq}


Substitute the value and solve the above expression

{eq}\begin{align*} {v_o} &= \sqrt {{{\left( {6.79} \right)}^2} + {{\left( {14.8} \right)}^2}} \\ &= \sqrt {46.1041 + 219.04} \\ &= \sqrt {265.1441} \\ &= 16.28\;{\rm{m/s}} \end{align*} {/eq}


Thus the magnitude of the glider's velocity is {eq}16.28\;{\rm{m/s}} {/eq}


Learn more about this topic:

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Linear Velocity: Definition & Formula

from MCAT Prep: Help and Review

Chapter 14 / Lesson 12
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