Physics, 10.09.2019 03:30, sharondacarruth1656

Astone is thrown vertically upward with a speed of 15.5 m/s from the edge of a cliff 75.0 m high .
(a) how much later does it reach the bottom of the cliff?
(b) what is its speed just before hitting?
(c) what total distance did it travel? aai8yoza helicopter is ascending vertically with a speed of 5.40 m/s. at a height of 105 m above the earth, a package is dropped from the helicopter. how much time does it take for the package to reach the ground?

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answered: englishwiz28

(a) 5.8s

(b) 41.36m/s

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 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;

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

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

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;

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

answered: Guest

answer: a. same unit

explanation:

answered: Guest

Not what goes around on earth but the universe too.

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