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Projectile Motion of a Batter Problem

  • #1
"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?
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
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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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