kinematic 2d motion, need help?

Question

a major-league pitcher can throw a baseball in excess of 37.0 m/s. if a ball is thrown horizontally at this speed, how much will it drop by the time it reaches a catcher who is 17.0m away from the point of release ?

no air resistence
my teacher told me you might need to use the quadratic formula.

if you can show your work, that will be great.

Answer 1

solve separately in two parts both vertically and horizontally
horizontally:
you know the inital speed (u) = 37m/s
you know the total distance (s) = 17m
you know the acceleration (a) = 0m/s² (no air resistance)

you can therefore work out the total time in motion with the equation:
s = ut + at²/2
17 = 37t
t = 17/37 seconds

now work out the vertical component
you know the the inital speed (u) = 0 [ball is thrown horizontally]
you know the total time taken (t) = 17/37
you know the acceleration (a) = -9.8m/s² [gravity]

therefore use the same equation to find the distance moved
s = -9.8/2 * (17/37)²
= -1.034404675 (9 s.f.)

:. ball will drop by 1.034404675m (9 s.f.)

Answer 2

The question you have given has the components
velocity along x=37m/s
acceleration along x=0
Initial velocity along y=0
acceleration=-gm/s^2=10m/s^2
There fore by taking components the body travells a distance of 17m along X-direction and then the time i=17/37s
Te distance it travells along y-direction is 1/2 g t^2 =1.055m

Answer 3

Horizontal motion x= v*t so 17 = 37 *t and t =0.46 s to reach te
catcher
vertical motion y = -1/2gt^2 as vertical speed at start is 0
drop = 4.9*(0.46)^2 =1.04 m

Answer 4

time = 17/37 = 0.02703 sec
y = 1/2 g t² = 1/2(9.81)( 0.02703)² = 0.0038 meter