A major league baseball player throws a hard ball (0.5 lbs) parallel to the grou

Question

A major league baseball player throws a hard ball (0.5 lbs) parallel to the ground for maximum speed towards home plate. The ball's speed right before the batter attempts to hit the ball is 90 miles per hour. If the batter hits a home run contacting the ball 3 feet above the ground and the ball reaches a maximum height of 103 feet above the warning track (330 ft. from home plate) after flying directly above second base, what is the impulse delivered by the bat to the ball?

Use 32.3 ft per second^2 for acceleration due to gravity. 5280 ft in 1 mile.

Explanation / Answer

u (initial velocity ) -= 90 miles per hour= 132 ft/sec

we can use conservation of energy to calcualte teh velocity imparted to basevall on hitting,

mgh = 1/mv^2

32.3 ( 100 ) = 1/2 v^2

v= 80.37 ft/sec

angle = tan^-1 ( 100/330 ) = 18.26 degree apprx

Vx = 80.37 cos 18.26 = 76.322 ft/sec

Vy = 80.37 sin 18.26 = 25.182 ft/sec

Ix) = 0.5 ( 132 + 76.322) = 104.161

Iy = 0.5 (25.182) = 12.591

I = sqroot ( 104.161^2 + 12.591^2) = 104. 9 pprx

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