a pitching machine throws a baseball at a batter with an initial velocity of 22.0 m/s at an angle of 30 degrees to the horizontal. the ball is 1.60 m above the ground when it is pitched.

a) if the batter hots the ball at the same height as it was launched how long is the ball in the air on its way to the batter?

b)how far is the ball from the pitching machine when it is hit?

c) if the batter misses the ball, where dose the ball land, relative to the pitching machine? relative to the batter?

4 answers

Vo = 22m/s @ 30o.
Xo = Hor. = 22*cos30 = 19.1 m/s.
Yo = Ver. = 22*sin30 = 11 m/s.

b. Range = Vo^2*sin(2A)/g.
Range = (22)^2*sin(60)/9.8 = 42.8 m. =
hor. distance.

a. Range = Xo * T = 42.8 m.
19.1 * T = 42.8
T = 2.25 s. = Time in air.

c. Tr = T/2 = 2.25/2=1.125 s.=Rise time.
hmax = ho + (Y^2-Yo^2)/2g.
hmax = 1.6 + (0-121)/-19.6 = 7.77 m.
Above gnd.

hmax = Vo*t + 0.5g*t^2 = 7.77 m.
0 + 4.9t^2 = 7.77
t^2 = 1.586
Tf = 1.26 s. = Fall time.

Dx=Xo*(Tr+Tf)=19.1*(1.125+1.26)=45.6 m.

do you go to Anderson CVI? :o

No, i've never heard of it.

I go to anderson :|