Question

1.On a spacecraft, two engines are turned on for 542 s at a moment when the velocity of the craft has x and y components of v_0x = 2960 m/s and v_0y = 5010 m/s. While the engines are firing, the craft undergoes a displacement that has components of x = 3.31 x 10⁶ m and y = 4.76 x 10⁶ m. Find the (a) x and (b) y components of the craft's acceleration.

2.A horizontal rifle is fired at a bull's-eye. The muzzle speed of the bullet is 690 m/s. The gun is pointed directly at the center of the bull's-eye, but the bullet strikes the target 0.028 m below the center. What is the horizontal distance between the end of the rifle and the bull's-eye?

3.A golf ball rolls off a horizontal cliff with an initial speed of 14.0 m/s. The ball falls a vertical distance of 12.7 m into a lake below.(a) How much time does the ball spend in the air? (b) What is the speed v of the ball just before it strikes the water?

4.A rocket is fired at a speed of 85.0 m/s from ground level, at an angle of 38.0

Answer 1

1

v= .5(t^2)+vo*t
3733=(5/2)*(568^2) + (v0x)*568
4895=(7.15/2)*(568^2) + (v0y)*568
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v0x=-1413.42
v0y=-2021.98
the negative sign means that your initial velocities are toward negative borders.
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to calculate total magnitude of velocity you can use this formula(cause forces are perpendicular):
vt=sqrt((v0x)^2 + (v0y)^2)
=> vt=2467 m/s

2

you are supposed to use the acceleration due to gravity (9.8 meters per second squared) to calculate the answer.

Figure out how long in seconds it would take for the bullet to fall that distance.

multiply the fall time of the bullet by 605 meters.

Remember:

D = 1/2aT^2

^2 is a way of saying squared if you can't get the superscript to go.

D = distance moved (downward)
a = acceleration due to gravity
T = time

T^2 = time squared, but let's call that Y for now.

so D = 0.05
a = 9.8 meters per second

0.05 = (0.5) x (9.8) Y .......I am using 0.5 because that is the same thing as 1/2 and is easier if you are using a calculator.

this is simple algebra. Remember once you solve for Y, you need to SQUARE ROOT it to get time.

Once you have Time, just multiply it by 605 to find how far away the target is

3

The time in the air is found with the equation D = Vit + 1/2a(t)squared
The vertical velocity initial is zero. The distance is down, so it is
negative(-15.0m). The acceleration is -9.8m/s2..
The Vit part of the equation is zero, so leave it out.
Therefore t = the squareroot of 2d/a or the squareroot of 2(-15)/-9.8
or 1.75s.

b. The vertical final velocity = Vi + at.
Vi = 0, so Vf = 9.8(1.75) = 17.15m/s.

The speed when the ball hits the ground is the hypotenus of a
triangle with 14.5m/s on one side and 17.15m/s on the other.
Find the squareroot of 14.5 squared plus 17.15 squared
or 22.46/s

4