A car drives off a vertical cliff that is 122.5m tall If the car lands 100m from the base of the cliff what was the cars speed as it drove off the cliff?

Answer 1

Equation: v=d/t, meaning velocity equals distance divided by the time interval.

Divide 54m by the five to get the time. Then divide 130m by the time you got to get the speed; also known as velocity.

Answers: Time= 10.8, Speed= 12.04m/s

I wonder where I got the 5 from...

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Figure out your knowns and unknown in the horizontal and vertical components.

x=130m, Vxo=?, ax=0, y=-54m, Vyo=0, ay=-9.8m/s2

Next, figure out the time it takes for the car to drive off the cliff using this equation for the Y component and solving for t:

y=y0+vyo+1/2gt2

y=0+0 +1/2(-9.8)t2

y=-1/2gt2, solving for t yields: t= sq root of 2y/-g. If all adds up well should equal 3.32s

Next, solve for the x components using this equation:

x=xo=vxo=1/2gt2

solving for vxo=x/t which equals 130 m/3.32s= 39.2 m/s

I hate physics.

Answer 2

first we need to get the time t = √ (2h/g ) = √ ( 2*122.5/9.8 = 5 s

distance = 100 m

therefore initial velocity is 100/5 = 20.

P.s: novelstars provides wrong choices. but you can choose A