A baseball player hits a home ball is hilt at a height of 1.0m above the ground.

Question

A baseball player hits a home ball is hilt at a height of 1.0m above the ground. What is the Initial speed of the balt (use g - 10 m/2) run, and the baseball just dear a wall 22 m located 120 m from the home plate. The ball is hit at an angle of 30 degrees to the horisontal, and air resistence is negigble. Assume the O-1.99E+03 m/s O 66.9 m/s O 5.35 m/s 44.6 m/s 140. m/s Submit Anewer Tries 0/3 How much time does it take for the bal to reach the wall? 3.42 3.11 s 6.21 3.11 0 9.66 Find the components of the velocity and the speed of the bail as it reaches the wa 77.2, 26.3,-39.6 [m/s) 38.6, -8.78, 39.6 [m/s) 42.5,-9.66, 83.2 (m/s) 239, -55.3, 273 (m/s) omit Answer Tries 0/ 74.6 [m) 24.9 (m] o (m] 12.4 [m] 49.7 (m]

Explanation / Answer

we have

Vi = initial speed of ball off bat, Vx = velocity x direction, Vy = velocity y direction
Dx = distance in x direction (120m); Dy = height in y direction (22m), Di = initial height (1m)
a = acceleration due to gravity (-9.81 m/s^2), t = time

Vy = Vi(sin30) = .5Vi
Vx = Vi(cos30) = .866Vi
Dx = Vx(t)
120 = .866Vi(t)
138.57 = Vi(t)
Dy = -1/2at^2 + Vy(t) + Di
22 = -4.905t^2 + .5Vi(t) + 1
21 = -4.905t^2 + .5(138.6)
21 - 69.3 = -4.905t^2
48.3/4.905 = t^2
t^2 = 9.85
t = 3.14 seconds

Vi(t) = 138.6
Vi(3.14) = 138.6
Vi = 44.1 m/s

Vy = -a(t) + .5(Vi)
Vy = -9.81(3.14) + .5(44.1)
Vy = -30.80 + 22.05
Vy = -8.75 m/s

Vx = .866Vi
Vx = .866(44.1)
Vx = 38.2 m/s

Vf = sqrt(Vx^2 + Vy^2)
Vf = sqrt(1459.24 + 76.56) = 39.19 m/s

Maximum height is given by

h= Vi2(sin30)2/ 2g = (44.1)2(0.25)/(19.6) = 24.8 m  

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