Physics
posted by Sarah on .
A batted baseball leaves the bat at an angle of 30.0 degrees about the horizontal and is caught by an outfilder 375 ft from home plate at the same height from which it left the bat. (a) what was the initial speed on the ball? (b) how high does the ball rise about the point where is struck the bat?

break up the initial speed into vertical and horizontal components.
Using the horizontal component, and distance, you know how long it was in the air
375=Vicos30*time
time=375/cos30
Now, knowing time in air..
hfinal=hinitia+visin30*time1/2 g t^2
or
vi*sin30=1/2 g t
put in for time what you got before, and then solve for vi. 
so for time i got 2431 s
and i used vi*sin30=1/2 g t
vi * .9880 = 1/2 (9.8) (2431) +.9880
= 11912.888 is this correct 
Hardly. Two thousand seconds in the air?
timeinair=375ft/Vi*.866=433/Vi
Notice I am working in English units.
Vi*1/2=1/2 (33ft/sec^2)433/Vi
vi= sqrt (33*433) ft/sec
check my thinking, and work.