What is the speed of a ball thrown from 1.2m above the ground if it travels 13.7m horizontally before hitting the ground

i need this done quick

The trick is to find the time.

If the ball did not travel horizontally at all and was just dropped, how long would it take to hit the ground?

Givens

vi = 0

d = 1.2 m

a = 9.8 m/s^2

Formula

d = vi * t + 1/2 * a ^ t^2

Solution

1.2 = 0*t + 1/2 9.8 * t^2

1.2 = 4.9 t^2

t^2 = 1.2 / 4.9

t^2 = 0.245

sqrt (t^2) = sqrt (.245)

t = 0.4949 seconds

Answer

You are not very clear about which answer you want. There are two of them. One is going horizontally and the other is vertically.

Vertical

vf^2 = vi^2 + 2*a*d

vi = 0

a = 9.8

t = 0.4949

vf^2 = 0^2 + 2 * 9.8 * 1.2

vf^2 = 23.52

sqrt (vf^2) = sqrt (23.52)

vf = 4.85 m/s

Horizontal

This is likely what you are looking for.

There is 0 acceleration horizontally.

d = 13.5 m

t = 0.4949

s = ?

s = d/t

s = 13.5/0.4949

s = 27.28

Note

You should note a couple of things.

The horizontal and vertical speeds are not the same. The time is used for both the horizontal and vertical speeds. The horizontal and vertical distances are quite different. Horizontal accelerations for these type of questions is generally 0.