Physics Help! Thanks! 2-Dimensional Motion Problem?

Question

I'm not sure how to go about answering this. Any help would be appreciated.

A ball is projected horizontally from the edge of a table that is 1.00m high, and it strikes the floor at a point 1.20m from the base of the table.
a.) What is the initial speed of the ball?
b.) How high is the ball above the floor when its velocity vector makes a 45.0 degree angle with the horizontal?

Answer 1

The equations of motion are

x(t)=vi*t
Vx(t)=vi

y(t)=h-.5*g*t^2
Vy(t)=-g*t

where vi is the initial speed of the ball, and h is the height of the table

a) Solve for t using
y(t)=0=1-.5*9.81*t^2
t=sqrt(2/9.81)

plug this into x(t)
1.2=vi*sqrt(2/9.81)
vi=1.2/sqrt(2/9.81)
vi=2.66 m/s

b) find t when magnitudes of Vx(t)=Vy(t)
2.66=g*t
t=2.66/9.81

plug this into y(t)
y(t)=1-.5*9.81*(2.66/9.81)^2
=0.639 m off the floor

j

Answer 2

The reaction above by employing way of odu83 is close, yet its without a doubt 4H/R = Tan(theta) extremely do no longer ignore that R = 2 v0^2 sin(theta) cos(theta)/g Then at the same time as you divide H/R, you get tan(theta)/4 = H/R. then you definately remedy and get that theta = arctan(4H/R)