Question

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?

Answer 1

Answer:
`\theta =32.08 ^(\circ)`

Step-by-step explanation:

Given

Ball launcher is mounted on car at angle of
`39^(\circ)`

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
`u_x=9.7cos39+2.2=7.53+2.2=9.73 m/s`

`u_y=9.7sin39=6.10 m/s`

therefore Appeared trajectory angle is

`tan\theta =(6.10)/(9.73)=0.627`

`\theta =32.08 ^(\circ)`