A Bowling ball is rolled down the alley with the constant velocity of 1.6 m/sec at an angle of 85° to the starting line, the position of the person throwing a bowling ball can be represented by the point (0,0). where is the ball after six seconds?

We have been given that a Bowling ball is rolled down the alley with the constant velocity of 1.6 m/sec at an angle of 85° to the starting line, the position of the person throwing a bowling ball can be represented by the point (0,0).

We are asked to find the position of ball after 6 seconds.

We will use horizontal position and vertical position formula to solve our given problem.

Horizontal position is given by formula: [tex]x=v\cdot \text{cos}(\theta)\cdot t[/tex] and vertical position is given by formula: [tex]y=v\cdot \text{sin}(\theta)\cdot t[/tex],where,

v = Velocity,

t = Time,

[tex]\theta[/tex] = Angle to the starting line.

For our given problem [tex]v=1.6[/tex], [tex]t=6[/tex] and [tex]\theta=85^{\circ}[/tex].

[tex]x=1.8\cdot \text{cos}(85^{\circ})\cdot 6[/tex]

[tex]x=10.8 \cdot 0.087155742748[/tex]

[tex]x=0.9412820216784\approx 0.94[/tex]

[tex]y=1.8\cdot \text{sin}(85^{\circ})\cdot 6[/tex]

[tex]y=10.8\cdot 0.996194698092[/tex]

[tex]y=10.7589027\approx 10.76[/tex]

Therefore, the position of ball after 6 seconds would be [tex](0.94,10.76)[/tex].

Answer:

(0.8,9.6)

Step-by-step explanation:

x=1.6*cos85* 6=0.8

y=1.6*sin85 *6=9.6

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