A girl dropped a stone from the top of a tower 45m tall and a boy projected another stone vertically upwards at 25m/s at the same time. Determine the time taken for the two stones to meet.
Question
A girl dropped a stone from the top of a tower 45m tall and a boy projected another stone vertically upwards at 25m/s at the same time. Determine the time taken for the two stones to meet.
Solution
To solve this problem, we need to set the distances traveled by both stones equal to each other and solve for time.
Let's denote:
- the time taken for the two stones to meet as t (in seconds),
- the height of the tower as H (in meters), which is 45m,
- the initial velocity of the stone thrown upwards as u (in m/s), which is 25m/s,
- the acceleration due to gravity as g (in m/s²), which is approximately 9.8m/s².
The stone dropped from the tower will only be influenced by gravity, so its distance traveled will be 1/2 * g * t².
The stone thrown upwards will initially move against gravity, then fall back down, so its distance traveled will be u*t - 1/2 * g * t².
Setting these two distances equal to each other gives us the equation:
1/2 * g * t² = u*t - 1/2 * g * t².
Solving this equation for t gives us:
t = 2u / g.
Substituting the given values into this equation gives us:
t = 2*25m/s / 9.8m/s² = 5.1 seconds.
So, the two stones will meet approximately 5.1 seconds after they are released.
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