Q-1: A projectile fired at 30 degrees with a velocity of 15 m/s would have an x-

Question

Q-1: A projectile fired at 30 degrees with a velocity of 15 m/s would have an x-velocity component of ________ m/s and a y-velocity component of ________ m/s.

Q-2: Calculate the components of the following projectile’s velocities:

a) v = 35 m/s = 15o vx =____ vy = ____

b) v = 35 m/s = 30o vx =____ vy = ____

c) v = 35 m/s = 45o vx =____ vy = ____

d) v = 35 m/s = 60o vx =_____ vy = ____

e) v = 35 m/s = 75o vx =_____ vy = ____

f) v = 35 m/s = 90o vx =_____ vy = ____

Q-3: We can reverse the process and combine the two components of velocity back into one velocity fired at an angle.

The magnitude of velocity is found using the Pythagorean Theorem with vx and vy as the legs of a right triangle. For instance, the velocity of a projectile with an x-component of 7.2 and a y-component of 4.8 is Radical 7.2^2 + 4.8^2 =8.7 m/s.

The angle above the horizontal is found using the inverse tangent (tan-1) of the legs vy/vx. For instance, the angle of the projectile described above would be tan^-1 (4.8/7.2)= 34o.

Calculate the velocity magnitude and angle of the projectiles listed below:

a) vx = 5.6 vy = 6.4 v =_____ = ______

b) vx = 2.8 vy = 4.9 v =_____ = ______

c) vx = 8.1 vy = -7.2   v =_____ = _____

d) vx = -1.3 vy = -5.2 v = ____ = _____

Explanation / Answer

1 )

given

= 30 degrees

velocity V = 15 m /s

the x - component is

Vox = V cos30

Vox = 15 X cos30

Vox = 12.99 m/s

the y - component is

Voy = V sin30

Voy = 15 X sin30

Voy = 7.5 m/s

2 )

a) v = 35 m/s = 15o

vx = 35 X cos15

= 33.80 m/s

vy = 35 X sin15

= 9.05 m/s

b) v = 35 m/s = 30o

vx = 35 X cos30

= 30.31 m/s

vy = 35 X sin30

= 17.5 m/s

c) v = 35 m/s = 45o

vx = 35 X cos45

= 24.74 m/s

vy = 35 X sin45

= 24.74 m/s

d) v = 35 m/s = 60o

vx = 35 X cos60

=17.5 m/s

vy = 35 X sin60

= 30.31 m/s

e) v = 35 m/s = 75o

vx = 35 X cos75

= 9.05 m/s

vy = 35 X sin 75

= 33.80 m/s

f) v = 35 m/s = 90o

vx = 35 cos 90

= 0 m/s

vy = 35 X sin90

= 35 m/s

3 )

a) vx = 5.6 vy = 6.4

v = ( 5.62 + 6.42 )1/2

v = 8.50 m/s

= tan-1( vy / vx )

= 48.810

b )

vx = 2.8 vy = 4.9

v = ( 2.82 + 4.92 )1/2

v = 5.64 m/s

= tan-1( vy / vx )

= 60.250

c )

vx = 8.1 vy = -7.2

v = ( 8.12 + -7.22 )1/2

v = 10.83 m/s

= tan-1( vy / vx )

= -41.60

d )

vx = -1.3 vy = -5.2

v = ( -1.32 + -5.22 )1/2

v = 5.36 m/s

= tan-1( vy / vx )

= 75.960

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