# Help with physics?

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

• ?
6 days ago

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

• ?
6 days ago

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)

• jeffrcal
6 days ago

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.

• ?
6 days ago

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

• donpat
6 days ago

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 <——————–

• Amy
6 days ago

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

• ?
6 days ago

1.5 seconds

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