Question

A stone is thrown vertically upward with a speed of 15.5 m/s from the edge of a cliff 75.0 m high .

Answer 1

a) 2.64 s

We can solve this part of the problem by using the following SUVAT equation:

$s=ut+\frac{1}{2}at^2$

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,

$-75.0 = 15.5 t -4.9t^2\\4.9t^2-15.5t-75.0 = 0$

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:

$v=u+at$

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:

$v = 15.5 +(-9.8)(2.64)=-10.4 m/s$

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:

$s=ut+\frac{1}{2}at^2$

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,

$-105 = 5.40 t -4.9t^2\\4.9t^2 -5.40 t-105=0$

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.

Answer 2

Answer:

(a) 5.8s

(b) 41.36m/s

(c) 99.52m

    5.21s

Explanation:

(a) This is the total time it takes the stone to reach its maximum height above the cliff and strike the ground at the base of the cliff after projection.

let the height attained by the stone above the cliff be $h_1$ and the time taken to attain this height be $t_1$. We can safely assume acceleration due to gravity to be taken as $g=9.8m/s^2$.

We use the first equation of motion under free fall to obtain $t_1$ as follows;

$v=u-gt_1............(1)$

given: u = 15.5m/s

Where $v$ is the final velocity and u is the initial velocity. The negative sign in the equation indicates the fact that the stone is moving upwards against gravitational pull. The final velocity $v=0$ at height $h_1$ because the stone will momentarily at the maximum height come to rest before it begins to fall back downwards.

Hence from equation (1) we obtain the following,

$0=15.5-9.8t_1\\9.8t_1=15.5\\hence\\t_1=15.5/9.8\\t_1=1.58s$

To get $h_1$ we use the third equation as follows;

$v^2=u^2-2gh_1$ ( the body is moving upward so g is negative)

$0^2=15.5^2-2*9.8*h_1\\0=240.25-19.6h_1\\19.6h_1=240.25\\therefore\\h_1=240.25/19.6\\h_1=12.26m$

Next we obtain the time it takes to fall back from the maximum height downwards to the base of the cliff. Let this time be $t_2$. We use the second equation of motion.

$H=ut+gt_2^2/2............(2)$

( g is positive because the stone is falling downwards)

However in this case, u = 0 because the stone is falling freely from rest downwards.

$H=h_1+75m=12.26+75\\H=87.26m$

Substituting into equation (2), we obtain;

$87.26=(0*t_2)+9.8t_2^2/2$

Simplifying further we obtain;

$4.9t_2^2=87.26\\t_2^2=87.26/4.9\\ =17.81\\t_2=\sqrt{17.81}=4.22s$

Hence the total time spent in air = 1.58+4.22 = 5.8s

(b) We use the third equation of motion to find the velocity with which the stone strikes the ground.

$v^2=u^2+2gH....... (3)$

the stone is falling downwards in this case from height H from rest, u = 0, v is the final velocity with which is strikes the ground. Equation (3) can therefore be reduced to the following form by putting u = 0;

$v=\sqrt{2gH}\\v=\sqrt{2*9.8*87.26} \\v=41.36m/s$

(c) The total distance travelled is given as follows;

$H_{total}=h_1+H\\H_{total}=12.26+87.26=99.52m$

When the package was dropped from the ascending helicopter, it will be projected upwards with an initial velocity equal to that of the helicopter, attain a maximum height and then fall back downwards. The total time spent in air by the package is the sum of the time it takes to attain maximum height and the time it takes to fall to the ground from the maximum height. This solution is similar to that of part (a) of this question.

To find the time it takes to attain maximum height, we use equation (1): v = 0, u = 5.4m/s and g is negative since the package is moving upward against gravity. Hence;

$0^2=5.4^2+9.8t_1\\9.8t_1=5.4\\t_1=5.4/9.8\\t_1=0.55s$

Similarly to the previous solution, we obtain the maximum height as follows;

$v^2=u^2-2gh_1\\0^2=5.4^2-2*9.8*h_1\\19.6h_1=29.16\\h_1=29.16/19.6\\h_1=1.49m$

therefore maximum height is

H = 105+1.49 = 106.49m

The time taken by the package to fall from H to the ground is given by equation (2), where u = 0 since the package is falling from rest; g is positive in this case.

$106.49=(0*t)+ 9.8t^2/2\\106.49=4.9t^2\\t^2=106.49/4.9=21.73\\t=\sqrt{21.73}=4.66s$

therefore the total time spent by the package before striking the ground is given by;

$t_{total}=0.55s+4.66s=5.21s$