bill and bob are both on the roof of a building 176.4 m high. bill drops a penny straight down at the same instant, bob throws a dime vertically upwards. the dime is thrown over the age when it comes back down it misses the roof and follows the penny to the street below. how fast was the dime thrown if when the, penny hits the ground the dime is still at a height of 58.8m?
To solve this problem, we can use the equations of motion for freely falling objects.
First, let's analyze the motion of the penny. The distance it falls vertically is equal to the height of the building, which is 176.4 m. We can use the equation:
s = ut + (1/2)at^2,
where s is the distance, u is the initial velocity, t is the time, and a is the acceleration. Since the penny is falling straight down, the initial velocity is 0 m/s and the acceleration due to gravity, a, is -9.8 m/s^2 (negative because it acts in the opposite direction to the upward direction we've defined). Plugging in the values:
176.4 = 0 + (1/2)(-9.8)t^2.
Simplifying the equation, we get:
176.4 = -4.9t^2.
Now, let's analyze the motion of the dime. It is thrown vertically upwards, reaches its maximum height, and then falls back down. When the penny hits the ground, the dime is still at a height of 58.8 m, which is the maximum height it reaches. We can use the equation:
v^2 = u^2 + 2as,
where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance. Since the dime starts from the ground, its initial velocity, u, is 0 m/s. The distance, s, is the difference between the maximum height and the original height, which is 176.4 - 58.8 = 117.6 m. Plugging in these values:
0 = u^2 + 2(-9.8)(117.6).
Simplifying the equation, we get:
u^2 = 2(9.8)(117.6).
Finally, we can solve for the initial velocity of the dime, u. Taking the square root of both sides of the equation:
u = sqrt(2(9.8)(117.6)).
Calculating the value, we find:
u ≈ 46.8 m/s.
Therefore, the dime was thrown with an initial velocity of approximately 46.8 m/s.