A stone is dropped from the top of a cliff. It hits the ground below 4.2 seconds after being dropped. How high is the cliff?
i allready know that Accleration-9.8 initial velocity is 0 and time is 4.2
but how do you find distance without final velocity
i allready know that Accleration-9.8 initial velocity is 0 and time is 4.2
but how do you find distance without final velocity
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y = y0 + v0*t + 0.5*a*t^2
y - final height
y0 - initial height
v0 - initial velocity
a - acceleration
t - time
We are solving for y0, y=0, v0 = 0, a = -9.8, t = 4.2
0 = y0 + 0.5*a*t^2
y0 = -0.5*a*t^2
y0 = -0.5*(-9.8)*(4.2^2)
y0 = 86.4m
y - final height
y0 - initial height
v0 - initial velocity
a - acceleration
t - time
We are solving for y0, y=0, v0 = 0, a = -9.8, t = 4.2
0 = y0 + 0.5*a*t^2
y0 = -0.5*a*t^2
y0 = -0.5*(-9.8)*(4.2^2)
y0 = 86.4m