An apple falls from a nearby tree and the instantaneous velocity of that apple is 15 m/s. How far has it fallen?
7 Answers

Since velocity , v = acceleration , a x time t = a x t , where a is 9.8 m/sec2 and v is 15 m /sec , so , time t in seconds = 15 m/sec / 9.8 m/sec2 = 1.530612245 seconds , so in this time t taken distance travelled ( falling height ) , Ht in meters = 1/2 a x t2 = 0.5 x 9.8 x ( 1.530612245 )2 = 11.47959184 meters or about 11.48 meters

LOOK AT YOUR ENGINEERING UNITS!
You know 2 engineering units: m/s^2 and m/s
You want to know: s
How do you set up an equation that will give you an answer with engineering units of “s”?
x s = 15 m/s / (9.8 m/s^2)
= (15/9.8) (m*s^2)/(s*m)

v = at so 15 = 9.8*t therefore t = 1.5 sec.
Now calculate the distance; since initial speed and displacement are both zero:
D = .5 (9.8)(1.5^2) = 11 meters.

h = V^2/2g = 15^2/19.6 = 11.5 m approx

You have :
—————
a = dv/dt = g ——> dt = dv/a
v = dy/dt —————> dt = dy/v
dt = dv/a = dy/v
v dv = a dy = g dy
v1^2 – v0^2 = ( 2 ) ( – 9.807 m/s^2 ) ( y1 – y0 )
y1 – y0 = [ ( 15 m/s )^2 ] / [ ( 2 ) ( – 9.807 m/s^2 ) ]
y1 – y0 = – 11.47 m
The apple has fallen 11.47 m when v equals 15 m/s <——————–
Please award Best Answer to the most correct and most helpful response to your question.

2ax = (v_f)^2 – (v_i)^2
2(9.81 m/s^2)x = (15 m/s)^2 – (0 m/s)^2
x = 11.47 m

1.5 seconds
☆ Sent from iOS Dr. Know! 1.0