A stone is dropped from a building 50 m high. Its acceleration at any time, t, can be approximated by the equation a = -10. How long does it take for the stone to hit the ground at the foot of the building?
Can someone please help me with this. Thanks in advance.
Answer: 3.16 s
distance = 1/2 * acceleration * time^2
-50 = 1/2 * -10 * t^2 ... 10 = t^2
h = ho - 0.5g*t^2 = 0.
50 - 5*t^2 = 0,
t =
To find the time it takes for the stone to hit the ground, you can use the following kinematic equation:
s = ut + (1/2)at²
Where:
s = distance (50 m)
u = initial velocity (0 m/s, since the stone is dropped)
a = acceleration (-10 m/s²)
t = time
Now, substitute the given values into the equation:
50 = 0 + (1/2)(-10)t²
Simplify and solve for t:
50 = -5t²
Divide both sides by -5:
-10 = t²
Take the square root of both sides:
t = √(-10)
Since time cannot be negative in this context, we discard the negative solution.
t ≈ 3.16 seconds
Therefore, it takes approximately 3.16 seconds for the stone to hit the ground at the foot of the building.
To find the time it takes for the stone to hit the ground, we can use the equation of motion:
s = ut + (1/2)at^2
Where:
- s is the distance traveled (in this case, the height of the building, which is 50 m)
- u is the initial velocity (which is 0 since the stone is dropped)
- a is the acceleration (which is -10 m/s^2)
- t is the time we are trying to find
Since the stone is dropped, the initial velocity (u) is 0. Therefore, the equation simplifies to:
s = (1/2)at^2
Plugging in the values we have:
50 = (1/2)(-10)t^2
Now, let's solve for t.
First, simplify the equation:
50 = -5t^2
Divide both sides of the equation by -5 to isolate t^2:
-10 = t^2
Take the square root of both sides of the equation:
t = √(-10)
Since the square root of a negative number is not a real number, we can conclude that the stone will not hit the ground in real time.
However, it seems there may be a mistake in the provided information or calculations. Please double-check the values to ensure accuracy.