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Olin [163]
3 years ago
12

A cannonball is tired with an initial speed of 40m/s and a launch angle of 30degrees from a cliff that is 25m tall. a. What is t

he flight time of the cannonball b. How far away does the cannonball land c. Using two graphs, plot the horizontal speed of the projectile versus time and the vertical speed of the projectile versus time (from the initial launch of the projectile to the instant i strikes the ground).

Physics
1 answer:
ella [17]3 years ago
7 0

Answer:

Part a)

t = 5.1 s

Part b)

x = 176.7 m

Explanation:

Initial speed of the launch is given as

v_i = 40 m/s

angle = 30 degree

Now the two components of the velocity

v_x = v_i cos30

v_x = 34.6 m/s

similarly we have

v_y = v_i sin30

v_y = 20 m/s

Part a)

Now to find the time of flight we will have

\Delta y = v_y t + \frac{1}{2}at^2

-25 = 20 t -\frac{1}{2}(9.81)t^2

t = 5.1 s

Part b)

Distance at which the ball will land is given as

x = v_x t

x = 34.6 (5.1)

x = 176.7 m

Part c)

Since x direction there is no acceleration so it will move with uniform speed while in y direction it will accelerate due to gravity

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3 years ago
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A dolphin in an aquatic show jumps straight up out of the water at a velocity of 15.0 m/s. (a) List the knowns in this problem.
astra-53 [7]

Answer:

a)

Y0 = 0 m

Vy0 = 15 m/s

ay = -9.81 m/s^2

b) 7.71 m

c) 3.06 s

Explanation:

The knowns are that the initial vertical speed (at t = 0 s) is 15 m/s upwards. Also at that time the dolphin is coming out of the water, so its initial position is 0 m. And since we can safely assume this happens in Earth, the acceleration is the acceleration of gravity, which is 9.81 m/s^2 pointing downwards

Y(0) = 0 m

Vy(0) = 15 m/s

ay = -9.81 m/s^2 (negative because it points down)

Since acceleration is constant we can use the equation for uniformly accelerated movement:

Y(t) = Y0 + Vy0 * t + 1/2 * a * t^2

To find the highest point we do the first time derivative (this is the speed:

V(t) = Vy0 + a * t

We equate this to zero

0 = Vy0 + a * t

0 = 15 - 9.81 * t

15 = 9.81 * t

t = 0.654 s

At this time it will have a height of:

Y(0.654) = 0 + 15 * 0.654 - 1/2 * 9.81 * 0.654^2 = 7.71 m

The doplhin jumps and falls back into the water, when it falls again it position will be 0 again. So we can equate the position to zero to find how long it was in the air knowing that it started the jump at t = 0s.

0 = Y0 + Vy0 * t + 1/2 * a * t^2

0 = 0 + 15 * t - 1/2 * 9.81 t^2

0 = 15 * t - 4.9 * t^2

0 = t * (15 - 4.9 * t)

t1 = 0 This is the moment it jumped into the air

0 = 15 - 4.9 * t2

15 = 4.9 * t2

t2 = 3.06 s This is the moment when it falls again.

3.06 - 0 = 3.06 s

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3 years ago
A sample of copper has a volume of 23.4 cm3 if the density of copper is 8.9 gcm3 what is the coppers mass?
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The answer is:  " 208 g " .
_____________________________________________
Explanation:
__________________________________________
The formula/ equation for density is:
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D = m / V  ;  That is,  "mass divided by volume" ;
 
Density is expressed as:
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                   "mass per unit volume";  in which the "mass" is expressed in units of "g" ("grams") ;  and the "unit volume" is expressed in units of:
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           {Note the exact equivalent:  1 cm³ = 1 mL }.
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     "m" refers to the "mass" , in units of "g" (grams), which is unknown; and we want to find this value;
                 
     "V" refers to the "volume", in units of "cm³ " ;
               which is:  "23.4 cm³ " (given);
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    Multiply each side of the equation by "V" ; 
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      V * D = m ;   ↔   m = V * D ;
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           m  =  V * D ;
 
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The answer is:  " 208 g " .
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