Given:
The given value is .
To find:
The value of the given expression by using the Binomial approximation.
Explanation:
We have,
It can be written as:
Therefore, the approximate value of the given expression is 1.0002.
A ball is thrown directly upwards. Is there a point of trajectory where the ball has zero acceleration
1 answer:
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<u>Answer:</u>
a) Speed of mailbag after 3 seconds = 32.4 m/s
b) Package is 44.1 meter below helicopter
c) If the helicopter was rising steadily at 3.00 m/s
Speed of mailbag after 3 seconds = 26.4 m/s
Package is 44.1 meter below helicopter
<u>Explanation:</u>
a) We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.
Initial velocity = 3 m/s, acceleration = 9.8 and time = 3 seconds.
v = 3+9.8*3 = 32.4 m/s
Speed of mailbag after 3 seconds = 32.4 m/s
b) We have equation of motion , , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.
Velocity of helicopter = 3 m/s, time taken = 3 seconds, acceleration = 0 .
Distance traveled by helicopter = 9 meter.
Velocity of package = 3 m/s, time taken = 3 seconds, acceleration = 9.8 .
Distance traveled by package = 53.1 meter.
So package is (53.1-9)meter below helicopter = 44.1 m
c) Initial velocity = -3 m/s, acceleration = 9.8 and time = 3 seconds.
v = -3+9.8*3 = 26.4 m/s
Speed of mailbag after 3 seconds = 26.4 m/s
Velocity of helicopter = -3 m/s, time taken = 3 seconds, acceleration = 0 .
Distance traveled by helicopter = 9 meter.
Velocity of package = -3 m/s, time taken = 3 seconds, acceleration = 9.8 .
Distance traveled by package = 35.1 meter.
So package is (35.1+9)meter below helicopter = 44.1 m
Answer: a) The cliff is 532.05m high
b) Her speed just before hitting the ground is 102.12 m/s
Explanation: To solve This, I'll use a sketch diagram, attached to this solution,
In 3seconds, the teacher heard the echo of her initial scream back. We can obtain the distance the teacher had fallen at the end of 3 seconds using the equations of motion,
Y1 = ut + 0.5g(t^2)
Since she's falling under the influence of gravity, her initial velocity, u = 0m/s, g = 9.8m/s2, t = 3s
Y1, distance she fell through in 3 seconds = 0.5×9.8(3^2) = 44.1m
Let the total height of the cliff be (44.1 + x); where is the remaining height of cliff that the teacher will fall through.
Using the equations of motion again, we can obtain distance travelled by the sound waves in 3s. sound waves travel with a constant speed of 340m/s, no acceleration,
Y2 = ut + 0.5g(t^2) where g = 0, u = 340m/s, t = 3seconds
Y2 = 340 × 3 = 1020m
But in 3 secs, the sound waves would have travelled through the total height of the cliff (44.1 + x) and back to the teacher's current height, x. That is, 1020 = 44.1 + x + x
x = 487.95m
So, total height of cliff = 44.1 + 487.95 = 532.05m
b) the speed of the teacher just before she hits the ground.
Using the equations of motion again,
(V^2) = (U^2) + 2gs
Where v is the final velocity to be calculated
U is the initial velocity = 0m/s
g is acceleration due to gravity = 9.8m/s2
S is the total height she fell through, that is, the height of the cliff = 532.05m
(V^2) = 0 + 2×9.8×532.05 = 10428.18
V = √(10428.18) = 102.12m/s
QED!
