Question

A small rock is thrown vertically upward with a speed of 17.0m/s from the edge of the roof of a 26.0m 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.
Part A
What is the speed of the rock just before it hits the street?
Express your answer with the appropriate units.
Part B
How much time elapses from when the rock is thrown until it hits the street?
Express your answer with the appropriate units.

Answer 1

Answer:

A) v = 28.3 m/s

B) t =  4.64 s

Explanation:

A)

• Assuming no other forces acting on the rock, since the accelerarion due to gravity close to the surface to the Earth can be taken as constant, we can use one of the kinematic equations in order to get first the maximum height (over the roof level) that the ball reaches:

        $v_{f}^{2} - v_{o}^{2} = 2* g* \Delta h (1)$

• Taking into account that at this point, the speed of the rock is just zero, this means vf=0 in (1), so replacing by the givens and solving for Δh, we get:

       $\Delta h = \frac{-v_{o} ^{2}}{2*g} = \frac{-(17.0m/s)^{2} }{2*(-9.8m/s2)} = 14.8 m (2)$

• So, we can use now the same equation, taking into account that the initial speed is zero (when it starts falling from the maximum height) and that the total vertical displacement is the distance between the roof level and the ground (26.0 m) plus the maximum height that we have just found in (2) , 14.8m:
• Δh = 26.0 m + 14. 8 m = 40.8 m (3)
• Replacing now in (1), we can solve for vf, as follows:

       $v_{f} =\sqrt{2*g*\Delta h} = \sqrt{2*9.8m/s2*40.8m} = 28.3 m/s (4)$

B)

• In order to find the total elapsed from when the rock is thrown until it hits the street, we can divide this time in two parts:
• 1) Time elapsed from the the rock is thrown, until it reaches to its maximum height, when vf =0
• 2) Time elapsed from this point until it hits the street, with vo=0.
• For the first part, we can simply use the definition of acceleration (g in this case), making vf =0, as follows:

       $v_{f} = v_{o} + a*\Delta t = v_{o} - g*\Delta t = 0 (5)$

• Replacing by the givens in (5) and solving for Δt, we get:

       $\Delta t = \frac{v_{o}}{g} = \frac{17.0m/s}{9.8m/s2} = 1.74 s (6)$

• For the second part, since we know the total vertical displacement from (3), and that vo = 0 since it starts to fall, we can use the kinematic equation for displacement, as follows:

       $\Delta h = \frac{1}{2} * g * t^{2} (7)$

• Replacing by the givens and solving for t in (7), we get:

       $t_{fall} =\sqrt{\frac{2*\Delta h}{g}} = \sqrt{\frac{2*40.8m}{9.8m/s2} } = 2.9 s (8)$

• So, total time is just the sum of (6) and (8):
• t = 2.9 s + 1.74 s = 4.64 s