Gliding at a constant horizontal velocity of 18.0 m/s at an altitude (height) of

Question

Gliding at a constant horizontal velocity of 18.0 m/s at an altitude (height) of 220.12 from the surface of the earth, you notice your favorite professor a certain distance away on the ground looking toward you. You use your radar to find out how far your professor is and decide to give them a yummy moment by launching an M&M candy at a velocity of 5.0 / at an angle of 23° with respect to the glider, so as to have the M&M enter your professor’s wide open awe-struck mouth which is at a height of 0.82m from the earth’s surface.

Explanation / Answer

y-component of initial velocity of candy, voy = 5*sin(23)

= 1.95 m/s

x-component of velocty of candy, vox = 18 + 5*cos(23)

= 22.6 m/s

Let t is the time taken for the candy to reach your professor.
Use kinemtaic euqation,

-h = voy*t - 0.5*g*t^2

-(220.12 - 0.82) = 1.95*t - 0.5*9.8*t^2

4.9*t^2 - 1.95*t - 219.3 = 0

on sloving the above equation,

we get, t = 6.9 s

so, displacement in x-direction, x = vox*t

= 22.6*6.9

= 155.94 m

displacement in y-direction, y = -(220.1 - 0.82)

= -219.3 m

net dispalcement, d = sqrt(x^2 + y^2)

= sqrt(155.94^2 + 219.3^2)

= 269.1 m <<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<----------Answer

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