Your favorite baseball player (Sammy Sosa? Mark McGwire? Dot Richardson?) has hit a home run in which the baseball ended up in the stands. The horizontal distance travelled by the ball was measured to be 137 m (450 ft), the initial angle of the ball was 50° from the horizontal. It took 5 seconds for the ball to reach the stands.

  a) What was the initial speed of the ball?
  b) What is the height h of the stands above the ground? (for simplicity, assume the ball started at ground level).
  c) What was the speed of the ball when it hit the stands?

Solution:

a) This is clearly a projectile motion problem. Let's choose the x-axis horizontal and y-axis vertical. Choose the origin as the location of the batter (the start of the ball's flight). Since we know the time, and the distance travelled in the x-direction, we can find the initial speed from
      x = x₀ + v_0xt
      x = 0 + v_0xt
      x = v₀cost
or
      v₀ = x/cost
      = (137 m)/(cos50°)(5 s) = 42.6 m/s

b) Now use the y-component equation:
      y = y₀ + v_0yt - (1/2)gt²
      = 0 + v₀sint - (1/2)gt²
      = (42.6 m/s)sin50°(5 s) - (1/2)(9.8 m/s²)(5 s)²
      = 40.7 m

c) Now use the equation for the velocity as a function of time:
      v_x = v_0x = v₀cos
       = (42.6 m/s)cos50° = 27.4 m/s
      v_y = v_0y - gt = v₀sin - gt
        = (42.6 m/s)sin50° - (9.8 m/s²)(5 s)
      = -16.4 m/s
now, we need to combine the components to get the speed:
      v = (v_x² + v_y²)^1/2 = 31.9 m/s

Note: This should have been very familiar: the method of solution is identical to that from problem 3-36 of the homework (level II problem).

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