Projectile Motion problem?

Question

This is from a test review and this is the only one I am having trouble with... If someone could please explain it to me, I'd be very grateful. :)

Babe Ruth hit a home run that was estimated to have landed 180m from home plate and to have reached a height of 21.3m. Find the initial velocity of the baseball and the angle it made with the horizontal.

We have to use a combination of any of these equations:
x=vt
final v = initial v +at
y = (initial v)t +.5a(t^2)
(final v)^2= (initial v)^2 + 2ay

a=9.81 m/s^2

And he gave us the answer to check ourselves, and it's 48 m/s at 25 degrees.

Thank you!

Answer 1

The time to apogee is half the flight time
v(t)=0 at apogee
0=v*sin(th)-9.81* t
therefore
t=v*sin(th)/9.81

so the total flight time is
2*v*sin(th)/9.81

x(t) at impact is 180 m
x(t)=v*cos(th)*t
so
180=v*cos(th)*2*v*sin(th)/9.81
or
883=v^2*cos(th)*sin(th)

also
y(t)=v*sin(th)*t-.5*9.81*t^2
since t=v*sin(th)/9.81 at apogee
21.3=.5*v^2*sin^2(th)/9.81
or
418=v^2*sin^2(th)
divide by
883=v^2*cos(th)*sin(th)
and
418/883=tan(th)
th=25 degrees
solving for v
=sqrt(418)/sin(25)
=48 m/s

j

Answer 2

I feel your pain! I just had a physics exam today on projectile motion and newtons second law.

I would love to help you, but my head hurts too bad still!

Answer 3

s_y= 21.3m
a= 9.81m/s^2
at max height velocity in the y direction is 0
vfy^2 =voy^2 +2as
voy = sqrt(2gs_y)
voy =20.44 m/s
s_y=0.5at^2 +voy*t
0=0.5at^2 +voy*t
t =4.167 s
s_x=180
s_x = vox*t
vox= s_x/t
vox=43.2 m/s
v= sqrt(vox^2+voy^2)
v=47.8 m/s
angle =arctan ( 20.44/43.2)
angle =25.3