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?
- JimLv 710 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
- Anonymous10 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.