Two children are playing on a 154-m-tall bridge. One child drops a rock (initial velocity zero) at
t = 0.
The other waits 1.1 s and then throws a rock downward with an initial speed
v0.
If the two rocks hit the ground at the same time, what is v0
The time of the first rock fall is
t1=sqrt(2h/g)=sqrt(2•154/9.8) =5.6 s.
The time of the second rock motion øs
t2=t1-1.1 =5.6-1.1=4.5 s.
h=vₒt+gt²/2.
vₒ=(h- gt²/2)/t= h/t-gt/2=154/4.5 -9.8•4.5/2 =34.22-22.05=12.17 m/s
To find the value of v0, we can use the kinematic equations of motion. Let's assume that the acceleration due to gravity is -9.8 m/s^2.
For the first child who drops the rock, we can use the equation for free fall:
y = y0 + v0t + (1/2)gt^2
Since the initial velocity is zero, the equation simplifies to:
y = (1/2)gt^2
For the second child who throws the rock downward, we can use the same equation but with a negative initial velocity:
y = y0 + v0t + (1/2)gt^2
Substituting the given values, we have:
154 = (1/2)(-9.8)(1.1)^2
154 = -5.39
However, we know that the height of the bridge is positive, so there must be an error in the calculation. It seems that we made a mistake when considering the negative sign in the equation.
Let's correct it by considering the downward direction as negative. Therefore, the equation becomes:
y = y0 + v0t - (1/2)gt^2
Substituting the values again, this time considering the downward direction as negative:
154 = (1/2)(-9.8)(1.1)^2
Solving this equation, we find:
v0 = -29.4 m/s
Since the velocity is negative, it means that the rock is thrown downward. Therefore, the answer is that v0 = -29.4 m/s.