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djverab [1.8K]
3 years ago
13

PLEASE HELP! I don't get it at all! Speed is one thing; distance is another. Where is the arrow you shoot up at 50m/s when it ru

ns out of speed?
Physics
2 answers:
LuckyWell [14K]3 years ago
4 0
I got you b, V(final)^2=V(initial+2acceleration*displacement
So this turns to (0m/s)^2=(50m/s)^2+2(9.8)(d) so just flip it all around to isolate d so you get
-(50m/s)^2/2(9.8) = d so you get roughly 12.7555 meters up
alina1380 [7]3 years ago
3 0
... Gravity makes it go 9.8 m/s slower each second.
    So it runs out of speed in (50/9.8) = 5.1 seconds after the shot.

... At the beginning of the shot, its speed is 50 m/s.
    When it runs out of speed, its speed is zero.
    Its AVERAGE speed during the whole 5.1 seconds is  25 m/s .

... Going straight up for 5.1 seconds, at an average speed of  25 m/s,
     it rises to an altitude of

                               (25 m/s) x (5.1 sec)  =  127.5 meters  .

(You said you have an answer of 125 meters.
Whoever worked out that one probably used 10 m/s²
for the acceleration of gravity, instead of the 9.8 that I used.) 


Of course, once the arrow "runs out of speed", it starts falling,
faster and faster, until it hits the ground.  You could calculate
how long it takes to fall, and how fast it's going when it hits the
ground.  But all of that is for another question.

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c

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Answer:

The maximum height reached by the two blocks is approximately 0.1147959 × v₀²

Explanation:

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By the Law of conservation of linear momentum, we have;

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Plugging in the values for the velocities gives;

3·m × v₀ + m × 0 = (3·m + m)·v₃ = 4·m·v₃

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\therefore v_3 = \dfrac{3}{4} \times v_0 = 0.75 \times v_0

The kinetic energy, K.E. of the combined blocks after the collision is given as follows;

K.E. = 1/2 × mass × v²

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The potential energy, P.E., gained by the two blocks at maximum height = The kinetic energy, K.E., of the two blocks before moving vertically upwards

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m = The mass of the object at the given height

g = The acceleration due to gravity

h = The height at which the object of mass, 'm', is located

Therefore, for h = The maximum height reached by the two blocks, we have;

P.E. = K.E.

m \cdot g \cdot h =  \dfrac{9}{8} \cdot m\cdot v_0^2

h = \dfrac{\dfrac{9}{8} \cdot m\cdot v_0^2}{m \cdot g }  = \dfrac{9}{8} \cdot \dfrac{ v_0^2}{ g }  =  \dfrac{9}{8} \cdot \dfrac{ v_0^2}{ 9.8} = \dfrac{42}{392} \cdot  v_0^2 \approx 0.1147959   \cdot  v_0^2

The maximum height reached by the two blocks, h ≈ 0.1147959·v₀².

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