Answer:
Approximately [tex]4.77\; {\rm s}[/tex].
Explanation:
Let [tex]u[/tex] denote the initial velocity of this apple. Let [tex]v[/tex] denote the velocity of the apple right before landing (final velocity.) Let [tex]x[/tex] denote the displacement of this apple (from the edge to the bottom.) Let [tex]a[/tex] denote the acceleration of this apple. Let [tex]t[/tex] denote the time it takes for the apple to land.
The acceleration of this apple is [tex]a = (-9.81)\; {\rm m\cdot s^{-2}}[/tex]. This value is negative since the apple is accelerating downwards.
It is given that the initial velocity of the apple was [tex]u = 15\; {\rm m\cdot s^{-1}}[/tex]. Note that unlike [tex]a[/tex], the value of [tex]u[/tex] is positive since the apple was initially travelling upwards.
The displacement of the apple would be [tex]x = (-40)\; {\rm m}[/tex]- equal to the height of the cliff in magnitude, but negative since the apple would land at a location below the edge.
Since the acceleration of this apple is a constant value, the SUVAT equation [tex]v^{2} - u^{2} = 2\, a\, x[/tex] will apply.
Rearrange this equation and solve for [tex]v[/tex] (velocity of apple right before landing):
[tex]\begin{aligned}v^{2} &= 2\, a\, x + u^{2}\end{aligned}[/tex].
Note that the apple will be travelling downward right before it lands. Therefore, the value of [tex]v[/tex] (velocity right before the apple lands) will be negative:
[tex]\begin{aligned}v &= -\sqrt{2\, a\, x + u^{2}}\end{aligned}[/tex].
Substitute in [tex]a = (-9.81)\; {\rm m\cdot s^{-2}}[/tex], [tex]x = (-40)\; {\rm m}[/tex], and [tex]u = 15\; {\rm m\cdot s^{-1}}[/tex]:
[tex]\begin{aligned}v &= -\sqrt{2\, a\, x + u^{2}} \\ &= -\sqrt{2\times (-9.81)\; {\rm m\cdot s^{-2}}\times (-40)\; {\rm m} + (15\; {\rm m\cdot s^{-1}})^{2}} \\ &\approx -31.78\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].
In other words, the velocity of the apple would have changed from [tex]u = 15\; {\rm m\cdot s^{-1}}[/tex] to [tex](-31.78)\; {\rm m\cdot s^{-1}}[/tex] during the flight. The velocity change would be:
[tex]\begin{aligned} \Delta v &= v - u \\ &\approx (-31.78)\; {\rm m\cdot s^{-1}} - 15\; {\rm m\cdot s^{-1}} \\ &= -46.78\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].
At a rate of [tex]a = (-9.81)\; {\rm m\cdot s^{-2}}[/tex], the time it takes to achieve such velocity change would be:
[tex]\begin{aligned} t &= \frac{\Delta v}{a} \\ &\approx \frac{(-46.78)\; {\rm m\cdot s^{-1}}}{(-9.81)\; {\rm m\cdot s^{-2}}} \\ &\approx 4.77\; {\rm s} \end{aligned}[/tex].
