A stone is thrown from the top of a building with an initial velocity of 20 m/s downward. The top of the building is 60 m above the ground. How much time elapses between the instant of release and the instant of impact with the ground
1 answer:
Answer:
t = 2 s
Explanation:
In order to find the time taken by the stone to fall from the top of the building to the ground we can use 2nd equation of motion. 2nd equation of motion is as follows:
s = Vit + (0.5)gt²
where,
t = time = ?
Vi = Initial Velocity = 20 m/s
s = height of building = 60 m
g = 9.8 m/s²
Therefore,
60 m = (20 m/s)t + (0.5)(9.8 m/s²)t²
4.9t² + 20t - 60 = 0
solving this quadratic equation we get:
t = -6.1 s (OR) t = 2 s
Since, the time cannot be negative in magnitude.
Therefore,
<u>t = 2 s </u>
You might be interested in
Hey /人 ◕ ‿‿ ◕ 人\
The answer is transmission
uses a series of gears to transmit power to facilitate changes in speed .
GLAD TO HELP
~~~ ╔͎═͓═͙╗
~~~ ╚̨̈́═̈́﴾ ̥̂˖̫˖̥ ̂ )
Answer:
detecting and indicating an electric current
I believe the answer is c
Answer:
i think it would be B, a large factory
Explanation:
Answer:
[Rn]7s2
Explanation: