A rock is thrown down with an initial speed of 10.0 m/s from a bridge to the water below. It takes 2.75 s for the rock to hit the water.
Find the speed of the rock as it hits the water
vf=vi+gt
vf=10+9.8*2.75
54.45
To find the speed of the rock as it hits the water, we can use the following kinematic equation:
v = u + gt
Where:
v = final velocity (speed at which the rock hits the water)
u = initial velocity (speed at which the rock was thrown down)
g = acceleration due to gravity (approximately 9.8 m/s^2, assuming no other forces acting on the rock)
t = time taken for the rock to hit the water
Given:
u = 10.0 m/s
t = 2.75 s
Plugging the given values into the equation, we get:
v = 10.0 m/s + (9.8 m/s^2 × 2.75 s)
Calculating the right side of the equation, we have:
v = 10.0 m/s + 26.95 m/s
Adding the values, we find:
v = 36.95 m/s
Therefore, the speed of the rock as it hits the water is approximately 36.95 m/s.
To find the speed of the rock as it hits the water, we can use the equation:
v = u + gt
Where:
v = final velocity of the rock
u = initial velocity of the rock
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time taken for the rock to hit the water
Given:
u = 10.0 m/s
g = 9.8 m/s^2
t = 2.75 s
Substituting the given values into the equation:
v = 10.0 m/s + (9.8 m/s^2) * (2.75 s)
Now, let's calculate the speed of the rock as it hits the water.