If an object is dropped from a tall building and hits the ground 3.0 s later, what is the velocity of the object when it hits the ground? (Remember that the velocity is downward)
i got -29m/s
is this correct?
correct: -9.8*3
I got -29.43m\s but that's not taking air resistance Into consideration.
To find the velocity of the object when it hits the ground, you can use the laws of motion. In this case, the object is dropped, so its initial velocity is 0 m/s.
The equation that relates velocity, time, and acceleration is:
v = u + at
Where:
v = final velocity
u = initial velocity
a = acceleration
t = time
Since the object is dropped, acceleration due to gravity is acting on the object, and it is directed downward. The acceleration due to gravity is approximately 9.8 m/s².
Plugging in the values:
u = 0 m/s (initial velocity)
a = 9.8 m/s² (acceleration due to gravity)
t = 3.0 s (time)
v = 0 + 9.8 * 3.0
v = 0 + 29.4
v ≈ 29.4 m/s
Since the velocity is downward, it will be negative. Therefore, the correct answer is -29.4 m/s. So, it seems like you made a calculation error. The correct velocity is -29.4 m/s, which means the object is moving downward at a speed of 29.4 m/s.