Anonymous
Anonymous asked in Science & MathematicsPhysics · 1 month ago

# Motion Along a Straight Line ?

Jack drops a stone from rest off of the top of a bridge that is 20.6 m above the ground. After the stone falls 7.0 m, Jill throws a second stone straight down. Both rocks hit the water at the exact same time. What was the initial velocity of Jill's rock? Assume upward is the positive direction and downward is negative. (Indicate the direction with the sign of your answer.)

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• When all motion is going downwards it would make sence so have this direction as positive, but if that's what you require then so be it.

-20.6 = (1/2)(-9.8)t²....using s = ut + (1/2)at²

so, t² = 41.2/9.8

i.e. t = 2.05 seconds to hit the ground

When the stone has travelled 7 metres we have,

-7 = (1/2)(-9.8)t²

so, t² = 14/9.8

i.e. t = 1.2 seconds

Hence, time taken to fall from 7 metres down to hit the ground is 2.05 - 1.2 = 0.85

Now, 0.85 seconds is the total time taken by Jill's rock to hit the ground

so, -20.6 = 0.85u - 4.9(0.85)²

i.e. -20.6 = 0.85u - 3.54025

=> u = -17.05975/0.85

so, u = -20.1 m/s....i.e. 20.1 m/s downwards

:)>

• lagging time t' = √ 2h/g = √ 14/9.806 = 1.195 sec

dropped falling mode

-20.6*2 = -9.806*t^2

t = √ 41.2/9.806 = 2.050 sec

thrown falling mode

-20.6 = -Vi*(2.050-1.195) -4.903*(2.050-1.195)^2

-20.6 = -Vi*0.855 -4.903*0.855^2

Vi = -(20.6-3.58)/0.855 ) = -20 m/sec

• Jack drops a stone from rest off of the top of a bridge that is 20.6 m above the ground. After the stone falls 7.0 m, Jill throws a second stone straight down. Both rocks hit the water at the exact same time. What was the initial velocity of Jill's rock? Assume upward is the positive direction and downward is negative. (Indicate the direction with the sign of your answer.)

how long does first rock take to fall 7.0 m

d = ½at² + v₀t + d₀

d = ½gt²

t² = 7.0/4.9

t = 1.195 sec

the second rock takes 1.195 s to travel 20.6 m

20.6 = 4.9(1.195)² + v₀1.195

v₀1.195 = 20.6 – 7.0 = 13.6

v₀ = 11.38 m/s

Source(s): billrussell42