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tester [92]
4 years ago
11

An attacker at the base of a castle wall 3.80 m high throws a rock straight up with speed 9.00 m/s from a height of 1.70 m above

the ground. a) will the rock reach the top? b) if so what is its speed at the top? if not, what initial speed must it have to reach the top? c) find the change in speed of a rock thrown straight down from the top of the wall at an initial speed of 7.4 m/s and moving between the same two points.d) does the change in speed of the downward moving rock agree with the magnitude of the speed change of the rock moving upward between the same elevations? e) why does this agree or not agree.
Physics
1 answer:
blagie [28]4 years ago
8 0

Answer:

First of all the attack is gonna die because its 2020 who throws a rock up over a wall to kill someone(cavemen) and

it's gonna be ↓

explanation:

An attacker at the base of a castle wall 3.60 m high throws a rock straight up with speed 8.00 m/s from a height of 1.70 m above the ground.

(a) Will the rock reach the top of the wall?

(b) If so, what is its speed at the top? If not, what initial speed must it have to reach the top?

(c) Find the change in speed of a rock thrown straight down from the top of the wall at an initial speed of 8.00 m/s and moving between the same two points.

(d) Does the change in speed of the downward-moving rock agree with the magnitude of the speed change of the rock moving upward between the same elevations? Explain physically why it does or does not agree.

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6 0
2 years ago
A stone is dropped into a river from a bridge 41.7 m above the water. Another stone is thrown vertically down 1.80 s after the f
hram777 [196]

Answer:

31.75 m/s

Explanation:

h = 41.7 m

Let the initial velocity of the second stone is u

Let the time taken to reach to the bottom by the first stone is t then the time taken by the second stone to reach the ground is t - 1.8.

For first stone:

Use second equation of motion

h=ut+\frac{1}{2}gt^2

Here, u = 0, g = 9.8 m/s^2 and t be the time and h = 41.7

So, 41.7= 0 + 0.5 x 9.8 x t^2

41.7 = 4.9 t^2

t = 2.92 s ..... (1)

For second stone:

Use second equation of motion

h=ut+\frac{1}{2}gt^2

Here, g = 9.8 m/s^2 and time taken is t - 1.8 = 2.92 - 1.8 = 1.12 s, h = 41.7 m and u be the initial velocity

h=u\left ( t-1.8 \right )+4.9\left ( t-1.8 \right )^2    .... (2)

By equation the equation (1) and (2), we get

41.7=1.12 u +4.9 \times 1.12^{2}

u = 31.75 m/s

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Answer:

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