Lucy's Stone: A Physics Challenge

[calculator20]

Lucy stands on the edge of a vertical cliff and throws a stone vertically upwards. The stone leaves her hand with speed 15m/s at the instant her hand is 80m above the surface of the sea. Air resistance is negligible and the acceleration of free fall is 10m/s/s.

First part is to calculate the maximum height reached which I've done and is 11m.

Second part is to find the time for the stone to reach the sea. I have found the correct answer of 5.8s eventually using the quadratic equation which is fine but quite tricky. This is where my question comes in.

The mark scheme also mentions as an alternative finding 4.3s as the time to fall 91m added to 1.5 for max height or finding 3 to return to hand and then 2.8 to fall 80 m. This is annoying me as I can't see where these answers come from? I assume the 1.5 comes from v/a = 15/10 and that it is doubled for 3 but how are they getting 4.3 and/or 2.8? I've tried every combination of motion equation and I'm sure I'm missing something obvious, please help!

 

[Borek]

At the top of the trajectory stone doesn't move, so it has to fall 91 m with acceleration g and no initial speed. That yields 4.3 s.

When the stone goes down, once it reaches the cliff level it has exactly the same speed, but opposite direction - so you have a free fall with v₀ = 15 m/s (do you know why?) and 80 meters to go. That yields 2.8 s.

Up and down in 3 s is just a free fall with initial speed going up. Going up takes exactly the same time it takes to go down.

 

[adjacent]

When the stone goes down, once it reaches the cliff level it has exactly the same speed...

Not the cliff level,the hand level(80 m)

 

[Borek]

Not the cliff level,the hand level(80 m)

Good point, lousy wording on my side.

 

[HalfLostAndSearching]

There are a few different ways to approach this problem and it looks like the mark scheme is providing alternative methods for finding the time for the stone to reach the sea.

One method is to use the equation h = ut + 1/2at^2, where h is the maximum height, u is the initial velocity, a is the acceleration, and t is the time. In this case, we know h = 11m, u = 15m/s, and a = -10m/s^2 (since the stone is moving upwards, acceleration is negative). Plugging in these values, we get 11 = 15t + 1/2(-10)t^2. This can be rearranged into a quadratic equation and solved to find t = 5.8s. This is the method you used and it is the most straightforward approach.

However, another method is to use the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. We know that the final velocity at the top of the stone's trajectory is 0m/s (since it stops moving before falling back down). We also know that the initial velocity is 15m/s and the acceleration is -10m/s^2. Using these values, we can solve for t to find that it takes 3 seconds for the stone to reach the top of its trajectory.

Then, we can use the same equation v = u + at to find the time for the stone to fall back down to Lucy's hand. In this case, the initial velocity is 0m/s (since the stone starts falling from rest) and the acceleration is 10m/s^2 (since the stone is now moving downwards). We also know that the final velocity when the stone reaches Lucy's hand is 15m/s (since it is moving at the same speed as when she threw it). Plugging in these values, we get 15 = 0 + 10t, which can be rearranged to t = 1.5s.

So, the total time for the stone to reach the sea is 3 seconds to reach the top of its trajectory, plus 1.5 seconds to fall back down to Lucy's hand, for a total of 4.5 seconds. This is where the 4.3 seconds in the mark scheme may be coming