Question

A batted baseball leaves the bat with a velocity of 41.4 m/s at an angle of 51.2°. If the ball just barely clears the 3.05 m (10 ft) center field wall,
a. How long was the ball in the air?
___s
b. What must the range be from home plate to the center field wall?
___m
c. How high did it rise above the point where the bat struck it?
___m
d. The final velocity is vf =
___m/s @ __°

Answer 1

Vo = 41.4m/s[51.2o].
Xo = 41.4*Cos51.2 = 25.9 m/s.
Yo = 41.4*sin51.2 = 32.3 m/s.

a. Y = Yo + g*Tr.
0 = 32.3 - 9.8Tr, Tr = 3.29 s. = Rise time.

h = Yo*Tr + 0.5g*Tr^2.
h = 32.3*3.29 - 4.9*3.29^2 = 53.2 m.

h = ho - 0.5g*Tf^2.
3.05 = 53.2 - 4.9Tf^2,
4.9Tf^2 = 53.2 - 3.05 = 50.18,
Tf = 3.2 s. = Fall time.

Tr+Tf = 3.29 + 3.20 = 6.49 s. = Time in air.

b. Range = Xo*(Tr+Tf) = 25.9*6.49 = 168 m.

c. h = 53.2 m.(Part a).

d. Y^2 = Yo^2 + 2g*h.
Y^2 = 32.3^2 + 19.6(53.2-3.05).
Y = 45 m/s.

V = Sqrt(Xo^2+Y^2) = Sqrt(25.9^2+45^2) = 51.9 m/s[60.1o].



























)

Answer 2

Correction:
d. y^2 = Yo^2 + 2g*h.
Y^2 = 0 + 19.6*(53.2-3.05) = 982.94, Y = 31.35 m/s.

V = Xo+Yi = 25.9 + 31.35i = 40.7 m/s.[50.4].