Projectile Motion of a Batter Problem

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SUMMARY

The projectile motion problem involves a batter hitting a ball at a height of 1.22 meters with a launch angle of 45 degrees, resulting in a horizontal range of 107 meters. To determine if the ball clears a 7.32-meter high fence located 97.5 meters away, one must analyze the vertical position of the ball at that distance. The initial velocity vector can be expressed as v cos(θ)i + v sin(θ)j, and the position vector at time t is given by v cos(θ)ti + (h + v sin(θ)t - (g/2)t²)j. The solution requires solving for the initial velocity v and checking the vertical height at the horizontal distance of 97.5 meters.

PREREQUISITES
  • Understanding of projectile motion principles
  • Familiarity with vector components in physics
  • Knowledge of kinematic equations
  • Basic algebra for solving equations
NEXT STEPS
  • Calculate the initial velocity required for a projectile to reach a specific range
  • Learn how to derive the equations of motion for projectile trajectories
  • Explore the effects of launch angle on projectile distance and height
  • Study the concept of maximum height in projectile motion
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Students studying physics, educators teaching projectile motion, and anyone interested in applying kinematic equations to real-world scenarios.

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"A batter hits a pitched ball when the center of the ball is 1.22m above the ground. The ball leaves the bat at an angle of 45 degrees with the ground. With that launch, the ball should have a horizontal range (returning to the launch level) of 107 m. (a) Does the ball clear a 7.32-m high fence that is 97.5 m horizontally from the launch point? (b) What is the distance between the top of the fence and the center of the ball when the ball reaches the fence?"

Is there a way I can find time? Do I have to find time for this?
 
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You should know that with any projectile problem like this, if the inital speed is v m/s, the angle is θ, and the initial height is h m, then the initial velocity vector is
v cosθi+ v sinθ j. The velocity vector at time t (in seconds after the "launch") is v cosθi+ (v sinθ-gt)j and the "position" vector at time t is
v cosθti+ (h+ v sinθt- (g/t)t2)j. The ball has range 107 m. I'm not at all clear what "returning to the level launch" means. I would assume that range was measured to the point where the ball hits the ground: where the j component of position is 0 but the height of the "launch" is 1.22 m!? Any way, you can set the vertical component of position to 0 (or 1.22?) and the horizontal component to 107 so that you have 2 equation to solve for the two unknown numbers v and t- except that you are right- you don't need to find t. If it is easier, go ahead and eliminate t from the two equations and just solve for v- that's what you need to know! Once you know a, put it into the equation for horizontal distance, with horizontal distance equal to 95.5 and solve for that t. Now put that t into the vertical distance equation, find the vertical height of the ball at that time and see if it is larger than 7.32 m.
 

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