# A ball is thrown horizontally from a 18 m -high building with a speed of 2.0 m/s .?

How far from the base of the building does the ball hit the ground?

### 2 Answers

- JimLv 711 months ago
y(t) = ½at² +v₀t + h₀. <<<< memorize this. it's used extensively!!!

y is the distance

a is the acceleration

t is time and must agree with the time units of the other factors

v₀ is the initial velocity, 'from rest' means v₀ = 0

h₀ is initial height or position

'a' s Earth Standard Gravity in most projectile flights (but not always!!!)

"Standard gravity" is, by definition, 9.80665 m/s² (about 32.1740 ft/s²).

y(t) = ½at² +v₀t + h₀ <<< sub in your values

0 = ½ * -9.80665* t² +(0)t + 18

½ * 9.80665* t² = 18

9.80665* t² = 36

t² = 3.670978366720542

t = 1.915979740686352 sec

Sigfigs: round to 2 digits

t = 1.9 s

- Anonymous11 months ago
How long does the ball take to fall to the ground?

18m = 1/2 * 9.81 m/sec^2 * time^2

horizontal distance = time * 2 m/sec.