A ball is thrown horizontally from the roof of a building 60 m tall and lands 50 m from the base?

Question

What was the ball's initial speed? in m/s

Answer 1

consider the x and y directions separately.

y direction:

initial velocity=0

acceleration=-9.8m/s^2 (acceleration due to gravity)

displacement= -60m (negative number because displacement is in negative y direction. ie. downward)

time= ?

find time:

s=vit+1/2at^2

(s=displacement, vi=initial velocity,a=acceleration, t=time)

-60=0t+1/2(-9.8)t^2

-60=-4.9t^2

-60/-4.9=t^2

t= 3.5 seconds

This is also the time for the x direction.

x direction:

initial velocity=?

acceleration= 0

displacement=50m

time=3.5seconds

find vi:

s=vit+1/2at^2

50=vi*3.5+1/2(0)t^2

50=3.5vi

vi=14.29m/s

:. initial velocity is 14.29m/s

Answer 2

Time to land to the ground = time to travel horizontally

y-axis:

y=.5*g*t^2

60m=.5*9.81m/s^2*t^2

solve for t

x-axis:

x=v*t

v=x/t

v=50/t

Answer 3

dx = Vx(t)

50 m = Vx(t)

dy = Viyt + (1/2)(g)(t^2)

60 m = (0)(t) + (1/2)(9.8m/sec^2)(t^2)

60 m = (4.9m/sec^2)(t^2)

t = sqrt[60 m/4.9 m/sec^2]

t = 3.5 sec

Therefore,  50 m = Vx(3.5sec)

                Vx = 50 m/3.5 sec

                Vx = 14.3 m/sec   ANSWER

Hope this helps.

teddy boy