A ball player hits a home run and the baseball just clears awall 8.0 m high loca

Question

A ball player hits a home run and the baseball just clears awall 8.0 m high located 105.0m from home plate. The ball is hit atan angle of 39.6 degrees to the horizontal, and air resistance innegligible. Assume the ball is 1.1m above the ground. What is the initial speed? How much time does it take to reach the wall? What is the speed when it reaches the wall? A ball player hits a home run and the baseball just clears awall 8.0 m high located 105.0m from home plate. The ball is hit atan angle of 39.6 degrees to the horizontal, and air resistance innegligible. Assume the ball is 1.1m above the ground. What is the initial speed? How much time does it take to reach the wall? What is the speed when it reaches the wall?

Explanation / Answer

The maximum height reached by the baseball is H = (u2 * sin2/2g) or u = (2H * g/sin2)1/2 = [(2H *g)1/2/sin] ---------(1) H = 8.0 m,g = 9.8 m/s2 and =39.6o The distance of the wall from homeplate is S = ut + (1/2)gt2 ----------(2) S = 105.0 m and the value of u is obtained from equation(1) Substituting and solving equation (2) we get the time taken bythe baseball to reach the wall The speed when it reaches the wall v = u + gt S = 105.0 m and the value of u is obtained from equation(1) Substituting and solving equation (2) we get the time taken bythe baseball to reach the wall The speed when it reaches the wall v = u + gt

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