a) 2.64 s
We can solve this part of the problem by using the following SUVAT equation:
![s=ut+\frac{1}{2}at^2](https://tex.z-dn.net/?f=s%3Dut%2B%5Cfrac%7B1%7D%7B2%7Dat%5E2)
where
s is the displacement of the stone
u is the initial velocity
t is the time
a is the acceleration
We must be careful to the signs of s, u and a. Taking upward as positive direction, we have:
- s (displacement) negative, since it is downward: so s = -75.0 m
- u (initial velocity) positive, since it is upward: +15.5 m/s
- a (acceleration) negative, since it is downward: so a= g = -9.8 m/s^2 (acceleration of gravity)
Substituting into the equation,
![-75.0 = 15.5 t -4.9t^2\\4.9t^2-15.5t-75.0 = 0](https://tex.z-dn.net/?f=-75.0%20%3D%2015.5%20t%20-4.9t%5E2%5C%5C4.9t%5E2-15.5t-75.0%20%3D%200)
Solving the equation, we have two solutions: t = -5.80 s and t = 2.84 s. Since the negative solution has no physical meaning, the stone reaches the bottom of the cliff 2.64 s later.
b) 10.4 m/s
The speed of the stone when it reaches the bottom of the cliff can be calculated by using the equation:
![v=u+at](https://tex.z-dn.net/?f=v%3Du%2Bat)
where again, we must be careful to the signs of the various quantities:
- u (initial velocity) positive, since it is upward: +15.5 m/s
- a (acceleration) negative, since it is downward: so a = g = -9.8 m/s^2
Substituting t = 2.64 s, we find the final velocity of the stone:
![v = 15.5 +(-9.8)(2.64)=-10.4 m/s](https://tex.z-dn.net/?f=v%20%3D%2015.5%20%2B%28-9.8%29%282.64%29%3D-10.4%20m%2Fs)
where the negative sign means that the velocity is downward: so the speed is 10.4 m/s.
c) 4.11 s
In this case, we can use again the equation:
![s=ut+\frac{1}{2}at^2](https://tex.z-dn.net/?f=s%3Dut%2B%5Cfrac%7B1%7D%7B2%7Dat%5E2)
where
s is the displacement of the package
u is the initial velocity
t is the time
a is the acceleration
We have:
s = -105 m (vertical displacement of the package, downward so negative)
u = +5.40 m/s (initial velocity of the package, which is the same as the helicopter, upward so positive)
a = g = -9.8 m/s^2
Substituting into the equation,
![-105 = 5.40 t -4.9t^2\\4.9t^2 -5.40 t-105=0](https://tex.z-dn.net/?f=-105%20%3D%205.40%20t%20-4.9t%5E2%5C%5C4.9t%5E2%20-5.40%20t-105%3D0)
Which gives two solutions: t = -5.21 s and t = 4.11 s. Again, we discard the first solution since it is negative, so the package reaches the ground after
t = 4.11 seconds.