A skateboard rolls off a horizontal ledge that is 1.12 m high, and lands 1.48 m from the base of the ledge. How much time was he in the air?
- MorningfoxLv 75 months ago
The landing distance is not a factor. Anything that drops 1.12 m on Earth, starting with zero vertical velocity and without air resistance, will take about 0.478 seconds to drop.
1.12 = 0.5 * g * 0.478^2
- billrussell42Lv 75 months ago
t = √(2h/g) = √(2•1.12/9.8) = 0.478 s
falling object starting from rest
h is height in meters, t is time falling in seconds,
g is acceleration of gravity, usually 9.8 m/s²
v is velocity in m/s
h = ½gt²
t = √(2h/g)
v = √(2gh)
h = v²/2g
v = gt