A small rock is thrown vertically upward with a speed of 17.0m/s from the edge of the roof of a 32.0m tall building. The rock doesn't hit the building on its way back down and lands in the street below. Air resistance can be neglected. What is the speed of the rock just before it hits the street? How much time elapses from when the rock is thrown until it hits the street?

Question

A small rock is thrown vertically upward with a speed of 17.0m/s from the edge of the roof of a 32.0m tall building. The rock doesn't hit the building on its way back down and lands in the street below. Air resistance can be neglected. What is the speed of the rock just before it hits the street? How much time elapses from when the rock is thrown until it hits the street?

Answer 1

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Answer 2

To find the speed of the rock just before it hits the street, we can use the kinematic equation for vertical motion:

v^2 = u^2 + 2as

where:
v = final velocity (which is what we are trying to find)
u = initial velocity (17.0 m/s upwards)
a = acceleration (which is -9.8 m/s^2, as the rock is accelerating downwards under the influence of gravity)
s = displacement (which is the height of the building, 32.0 m in this case)

Now, let's substitute the values into the equation:

v^2 = (17.0 m/s)^2 + 2(-9.8 m/s^2)(32.0 m)

v^2 = 289.0 m^2/s^2 - 627.2 m^2/s^2

v^2 = -338.2 m^2/s^2

Since velocity cannot be negative, we take the positive square root of 338.2 to get the final velocity:

v ≈ 18.4 m/s

Therefore, the speed of the rock just before it hits the street is approximately 18.4 m/s.

To find the time elapsed from when the rock is thrown until it hits the street, we can use another kinematic equation for vertical motion:

s = ut + 1/2at^2

where:
s = displacement (32.0 m in this case)
u = initial velocity (17.0 m/s upwards)
a = acceleration (-9.8 m/s^2)
t = time (what we are trying to find)

Substituting the known values:

32.0 m = (17.0 m/s)t + 1/2(-9.8 m/s^2)t^2

32.0 m = 17.0 m/s * t - 4.9 m/s^2 * t^2

Rearranging the equation, we get a quadratic equation:

4.9 t^2 - 17.0 t + 32.0 = 0

To solve this equation, we can use the quadratic formula:

t = (-b ± √(b^2 - 4ac))/(2a)

In this case, a = 4.9, b = -17.0, and c = 32.0.

t = (-(-17.0) ± √((-17.0)^2 - 4(4.9)(32.0)))/(2(4.9))

t = (17.0 ± √(289.0 - 627.2))/(9.8)

t = (17.0 ± √(-338.2))/(9.8)

Since the value inside the square root is negative, there are no real solutions for time. This means that the rock does not hit the street within the time frame of the problem, so we can assume that the time is undefined in this case.

In summary, the speed of the rock just before it hits the street is approximately 18.4 m/s, and the time elapsed from when the rock is thrown until it hits the street is undefined.