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 will use the principles of motion and gravity.
First, let's consider the motion of the penny. It is dropped straight down, so its acceleration due to gravity is -9.8 m/s² (negative because it acts in the opposite direction to the positive direction we choose). We need to find the time it takes for the penny to hit the ground.
Using the equation of motion:
h = ut + (1/2)gt²
Where:
h = height (176.4 m)
u = initial velocity (0 m/s, as the penny is dropped)
g = acceleration due to gravity (-9.8 m/s²)
t = time
Plugging in the given values:
176.4 = 0 + (1/2)(-9.8)t²
Simplifying the equation:
176.4 = -4.9t²
Now solve for t:
t² = 176.4 / -4.9
t² = -36
t = ±√(-36)
Since time cannot be negative, we discard the negative solution:
t = √36
t = 6 seconds
So, it takes 6 seconds for the penny to hit the ground.
Next, let's consider the motion of the dime. It is thrown upwards by Bob and comes back down, missing the roof, and eventually hitting the ground at the same time as the penny. We need to find the initial velocity at which the dime was thrown.
Using the equation of motion:
h = ut + (1/2)gt²
Where:
h = height (58.8 m)
u = initial velocity (unknown)
g = acceleration due to gravity (-9.8 m/s²)
t = time (6 seconds)
Plugging in the given values:
58.8 = u * 6 + (1/2)(-9.8)(6)²
Simplifying the equation:
58.8 = 6u - 176.4
Rearranging the equation:
6u = 58.8 + 176.4
6u = 235.2
Dividing both sides by 6 to solve for u:
u = 235.2 / 6
u = 39.2 m/s
Therefore, the dime was thrown with an initial velocity of 39.2 m/s.