A falling stone accelerates at a constant rate of 10m/s.It is dropped from rest down a deep well and 3 sec later a splash is heard as it hits the water below.How fast will it be moving as it hits the water?
To find out how fast the stone will be moving when it hits the water, you can use the equation of motion for uniformly accelerated motion:
v = u + at
Where:
v is the final velocity
u is the initial velocity
a is the acceleration
t is the time taken
In this case, the stone is being dropped from rest, so the initial velocity (u) is 0 m/s. The acceleration (a) is given as 10 m/s². The time taken (t) is 3 seconds.
Now, we can substitute these values into the equation:
v = 0 + (10)(3)
v = 0 + 30
v = 30 m/s
Therefore, the stone will be moving at a speed of 30 m/s when it hits the water.