A ball is thrown horizontally from a height of 15.11 m and hits the ground with a speed that is 4.0 times its

Please need urgent help........ grantee 10 points.

the final speed = Sqrt[v hor^2 + v vert^2]

where vhor and vvert are the final horizontal and vertical velocities

since there are no horizontal forces acting, we know the horizontal velocity is the same throughout the flight, so the initial speed = horizontal speed;


and we are also told that final speed = 4 v0 where v0 is the initial speed

we can find the time of flight and final vertical velocity from the height of the building:

knowing the building is 15.11m, we know that

height = 1/2 gt^2 or t=sqrt[2h/g]=1.76s = time of flight

the vertical velocity increases at the rate of 9.8m/s/s, so the final vertical velocity is

9.8m/s/s x 1.76s = 17.2m/s


and we have:

final speed = 4 v0

final speed^2 = 16 v0^2

but final speed ^2 also = vhor^2 + vvert^2

but the horizontal speed is the same throughout, so v0=vhor and we get

vhor^2 + vvert^2 = 16 vhor^2

v vert^2 = 15 v hor^2

17.2^2 = 15 vhor^2

vhor = 4.4m/s