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At time ,,t=0, a tennis ball leaves the origin at 35m/s and 30 ° above the x axis. if gravity acts in the y direction, what is the ball's position at time = 2 s ?

User Spechter
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Final answer:

At t=2s, using the components of the initial velocity, the horizontal position (x) is calculated to be 60.621m, and the vertical position (y) is 15.4m from the origin.

Step-by-step explanation:

To find the ball's position at time t=2s after being launched at an initial velocity of 35m/s at a 30° angle above the x-axis, we need to calculate the horizontal (x) and vertical (y) components of the motion separately, assuming acceleration due to gravity (g) is 9.8 m/s² acting in the negative y direction.

To find the x-position, use the horizontal component of the velocity (vx = v0∙cos(θ)), which is unaffected by gravity:

  • vx = 35m/s ∙ cos(30°)
  • x = vx ∙ t
  • x = (35m/s ∙ cos(30°)) ∙ 2s
  • x = (35m/s ∙ √3/2) ∙ 2s
  • x = 60.621m

To find the y-position, use the vertical component of the velocity (vy = v0∙sin(θ)) and factor in the acceleration due to gravity:

  • vy = 35m/s ∙ sin(30°)
  • y = vy ∙ t - ½gt²
  • y = (35m/s ∙ 0.5) ∙ 2s - ½(9.8 m/s²)(2s)²
  • y = 35m - 19.6m
  • y = 15.4m

Thus, at time t = 2s, the tennis ball's position is x = 60.621m and y = 15.4m from the origin.

User Mahdikmg
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