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A baseball is hit by a bat and given a velocity of 40.0m/s at an angle of 30 degree above the horizontal. The height of the ball above the ground upon impact with the bat is 1.0 m  a) what maximum height above the ground does the wall reached b) a feilder is 110.0 m from home plate with the ball is hit and the balls trajectory is directly at home if he begins running at the moment the ball is hit and catches the ball. when it is still 3.0 m above the ground. How long does he run before catching the ball. C) how fast does he have to run in order to catch the ball
Asked by lonese3325 | 12 Jun, 2022, 01:42: PM
Part (a)

Maximum height reached by ball above ground  , hmax = ( u2 sin2α ) / ( 2 g )+ 1 m

where u = 40 m/s is initial  projection velocity , α is projection angle and g is acceleration due to gravity

hmax = [ ( 40 × 40 sin230 ) / ( 2 × 9.8 ) ] + 1 = 21.41 m

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Part (b)

Time taken by the ball to reach 3m above ground is determined from the following equation

h = ( u sinα ) t - [ (1/2) g t2 ]

in above equation h = 2 m , because initially the ball is hit at the height of 1 m above ground

2 = ( 40 sin30 t ) - ( 4.9 t2 )

From above quadratic equation we get t = 3.98 s ≈ 4 s

hence fielder runs for 4 s to catch the ball

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Part (c)

Speed of fielder = distance / time = 110 m / 4 s = 27.5 s
Answered by Thiyagarajan K | 12 Jun, 2022, 02:06: PM

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