# Ten Kinematic Problems Grade 12 Physics

#### Steb

1. Homework Statement
See attachment, it is the question sheet.

2. Homework Equations

d=vit+1/2at^2
vf=vi+at
d=vit-1/2at^2
d=vavt

3. The Attempt at a Solution
1. An object such as a ball let go by a person's hand falling vertically downward toward earth has a velocity and acceleration vector in the same direction. An object such as a cannonball launched vertically upward will have a velocity vector upward and an acceleration vector down due to gravity which means the velocity and acceleration vectors are in opposite directions momentarily.

2. Both bullets hit the ground at the same time because both objects have the same initial velocity of zero and acceleration of gravity in the y-component. The x-component does not affect the y-component which is why it is ignored.

3. See attachment with no title.
4. The stone thrown downward will hit the ground first. Both stones hit the ground at the same speed.
5. 15^2+35^2=r^2
38m=r
The magnitude of the football's displacement is 38m.
6. vf=vi+at
vf=0-9.8(2.5)
vf=-24.5m/s
The velocity of the fish is 25m/s down at 2.5 seconds after being dropped.
b)d=vit-1/2at^2
d=0+4.9(2.5)^2
d=11.15m
The pelican was 11.15m above the water when it dropped the fish.
7. See attachment
Not sure how to do 8, 9 or 10.

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#### Delphi51

Homework Helper
Looking good on the first few. In #5, you didn't find the direction. In #6, you failed to use the initial upward velocity and in part (b) the time of fall was 10 s, not 2.5 s.
I didn't find you solution for #7.
#8 is a 2D vector problem. Sketch the velocity vectors for the wind and airspeed (placed head to tail). Their total must be in the eastward direction. From the sketch you could find the unknown angle with the Laws of sines and cosines or by splitting the vectors into horizontal and vertical parts.

#9 is similar - to velocity diagrams.

#### WillemBouwer

Yup, all seems good... as delphi51 mentioned, remember to give a direction for the displacement vector in 5.
In 6, what is your argumant for using vi as 0? Do you not think it is travelling at the same speed as the pelican when released? rethink this one.
in #7 you can break up the components of in xy-axis format. The car is in linear motion up to the edge of the cliff so what parameter can be calculated for the vehicle at this point(end of cliff). take this parameter and break it into its x and y components and find the horizontal displacement seeing as you have the y displacement and use pathagoras to get the relative position. from here time could be calculated as well as final velocity, remember to give an angle to this as it is a vector. keep in mind, what is the vertical and horizontal acceleration of the vehicle after it left the cliff?
8 and 9 is as delphi51 stated.
#10... According to the projectile equations of motion:
Vi = sqrt((R^2g)/(Rsin2θ+2hcos^2θ))
where R is your launch range, h is the drop height from start point and θ is your launch angle which you have...

Hope you come right...

#### Steb

Looking good on the first few. In #5, you didn't find the direction. In #6, you failed to use the initial upward velocity and in part (b) the time of fall was 10 s, not 2.5 s.
I didn't find you solution for #7.
#8 is a 2D vector problem. Sketch the velocity vectors for the wind and airspeed (placed head to tail). Their total must be in the eastward direction. From the sketch you could find the unknown angle with the Laws of sines and cosines or by splitting the vectors into horizontal and vertical parts.

#9 is similar - to velocity diagrams.
Thanks Delphi51

#### Steb

Yup, all seems good... as delphi51 mentioned, remember to give a direction for the displacement vector in 5.
In 6, what is your argumant for using vi as 0? Do you not think it is travelling at the same speed as the pelican when released? rethink this one.
in #7 you can break up the components of in xy-axis format. The car is in linear motion up to the edge of the cliff so what parameter can be calculated for the vehicle at this point(end of cliff). take this parameter and break it into its x and y components and find the horizontal displacement seeing as you have the y displacement and use pathagoras to get the relative position. from here time could be calculated as well as final velocity, remember to give an angle to this as it is a vector. keep in mind, what is the vertical and horizontal acceleration of the vehicle after it left the cliff?
8 and 9 is as delphi51 stated.
#10... According to the projectile equations of motion:
Vi = sqrt((R^2g)/(Rsin2θ+2hcos^2θ))
where R is your launch range, h is the drop height from start point and θ is your launch angle which you have...

Hope you come right...
Thanks WillemBouwer