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You mount a ball launcher to a car. It points toward the back of the car 39 degrees above the horizontal and launches balloons at 9.7 m/s. When you drive the car straight ahead at 2.2 m/s, what is the initial angle in degrees of the ball's trajectory as seen by someone standing on the ground?

Respuesta :

Answer:[tex]\theta =32.08 ^{\circ}[/tex]

Explanation:

Given

Ball launcher is mounted on car at angle of [tex]39^{\circ}[/tex]

launching velocity is u=9.7 m/s

Speed of car =2.2 m/s

So in horizontal component of balloon speed of car will be added

Thus [tex]u_x=9.7cos39+2.2=7.53+2.2=9.73 m/s[/tex]

[tex]u_y=9.7sin39=6.10 m/s[/tex]

therefore Appeared trajectory angle is

[tex]tan\theta =\frac{6.10}{9.73}=0.627[/tex]

[tex]\theta =32.08 ^{\circ}[/tex]

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