Physics, 01.06.2021 23:30 kktiger14

Jack drops a stone from rest off of the top of a bridge that is 21.0 m above the ground. After the stone falls 7.0 m, Jill throws a second stone straight down. Both rocks hit the water at the exact same time. What was the initial velocity of Jill's rock

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Answer from: wshemekia3oyliv4

v = 19 m/s

Explanation:

Since both rocks hit the water at the same time, and we know the total height of the bridge, we can get the time that it takes to the first rock to travel this height, due it is said that it was dropped, using the following kinematic equation:

$\Delta h = \frac{1}{2} * g * t^{2} (1)$

Replacing by the givens (Δh =21 m and g = 9.8 m/s2), and solving for t we get:

$t = \sqrt{\frac{2*21m}{9.8m/s2}} = 2.1 s (2)$

Now we need the time traveled by the first rock when it fell 7 m, that when the second rock was thrown, because the difference between the total time and this one will be the flight time for the second rock.We can use the same equation (1) replacing Δh by 7 m, and solving for t, in (2) as follows:

$t = \sqrt{\frac{2*7m}{9.8m/s2}} = 1.2 s (3)$

So, the total flight time for the second rock it was just the difference between (2) and (3):

$t_{2} = t_{tot} - t_{1} = 2.1 s - 1.2 s = 0.9 s (4)$

Since we know the total distance traveled, and the time of flight, we can use the same kinematic equation than in (1) but now taking into account the initial speed for the second rock, as follows: $\Delta h = v_{o}* t_{2} + \frac{1}{2}* g*t_{2} ^{2} (5)$ Since we know the value of Δh = 21 m and t₂ =0.9s, replacing in (5) and solving for v₀, we get:

$v_{o} =\frac{(\Delta h -\frac{1}{2}*g*t^{2})}{t_{2} } = \frac{(21m-(\frac{1}{2}*9.8m/s2*(0.9s)^{2}) ) }{(0.9m)} = 19.0 m/s (6)$

which was the initial velocity of Jill's rock taking as positive the down ward direction.

Answer from: Quest

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6 $\frac{m}{s^{2} }$

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s

Answer from: Quest

note: the given question is incomplete.

complete question is given below.

which is an example of kinetic energy?

a. the energy stored in ethanol

b. a hockey puck sliding across the ice

c. a compressed spring

d. a ball sitting at the top of a ramp

solution: the kinetic energy is the energy posses by the virtue of the motion of the object.

in the given options, a, c and d are the example of potential energy because it is the energy posses by the virtue of the rest of the object. but option b has the kinetic energy as the object is in motion.

hence, a hockey puck sliding across the ice is the apex example of kinetic energy.

so, the option (b) a hockey puck sliding across the ice is the correct option.

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Jack drops a stone from rest off of the top of a bridge that is 21.0 m above the ground. After the s...