An inflatable raft (unoccupied) floats down a river at an approximately constant speed of 5.3 m/s. a child on a bridge, 60 m above the river, sees the raft in the river below and attempts to drop a small stone onto the raft. the child releases the stone from rest. in order for the stone to hit the raft, what must be the horizontal distance between the raft and the bridge when the child releases the stone?

Distance = 18.55 m

Step-by-step explanation:

Here, Firstly we have to find time taken to fall stone by the child from bridge.

Given, height(s) = 60 m

Thus using formula,

60 = 0 + 0.5 × 9.8 × t²

⇒ t² = 12.245

⇒ t = 3.5 sec

Now, we know speed of raft and time to reach below bridge.

We have, Speed = 5.3

t = 3.5 sec

Using Formula,

Speed = Distance ÷ Time

⇒ 5.3 = Distance ÷ 3.5

⇒ Distance = 5.3 × 3.5

⇒ Distance = 18.55 m

Thus, the horizontal distance between the raft and the bridge when the child releases the stone is 18.55 m.

are you sure this is middle schools math? ?

step-by-step explanation:

x^2 -12x + 36 + y^2 = 36

x^2 + y^2 = 12x

r^2 = 12rcos(theta)

r = 12 cos(theta)