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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?

User Mikewaters
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2 Answers

0 votes

Answer:

(0.8,9.6)

Explanation:

x=1.6*cos85* 6=0.8

y=1.6*sin85 *6=9.6

User CrazyPyro
by
8.8k points
2 votes

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

v = Velocity,

t = Time,


\theta = Angle to the starting line.

For our given problem
v=1.6,
t=6 and
\theta=85^(\circ).


x=1.8\cdot \text{cos}(85^(\circ))\cdot 6


x=10.8 \cdot 0.087155742748


x=0.9412820216784\approx 0.94


y=1.8\cdot \text{sin}(85^(\circ))\cdot 6


y=10.8\cdot 0.996194698092


y=10.7589027\approx 10.76

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

User Tom Verhoeff
by
8.2k points
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