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pilar22 · May 23, 2008, 10:58 PM

A football is thrown with an initial upward velocity component of 15.0 and a horizontal velocity component of 18.0 .

How high does it get above the ground?
How much time after it is thrown does it take to return to its original height?
How far has the football traveled horizontally from its original position?

pilar22 · May 23, 2008, 11:01 PM

A tennis player hits a ball at ground level, giving it an initial velocity of 28.0 at 52.0 above the horizontal.

What are the horizontal and vertical components of the ball's initial velocity?
How high above the ground does the ball go?
How long does it take the ball to reach its maximum height?
What is the ball's velocity at its highest point?
What is the ball's acceleration at its highest point?
For how long a time is the ball in the air?
When this ball lands on the court, how far is it from the place where it was hit?

pilar22 · May 23, 2008, 11:03 PM

A man stands on the roof of a 10.0 -tall building and throws a rock with a velocity of magnitude 24.0 at an angle of 34.0 above the horizontal. You can ignore air resistance.

Calculate the maximum height above the roof reached by the rock.
Calculate the magnitude of the velocity of the rock just before it strikes the ground.
Calculate the horizontal distance from the base of the building to the point where the rock strikes the ground.

pilar22 · May 23, 2008, 11:03 PM

You swing a 3.00 stone in a circle, using the full length of a thin 65.0 rope.

At what speed should you swing it so its acceleration will be 9.8 ?

pilar22 · May 23, 2008, 11:06 PM

A baseball thrown at an angle of 55.0 above the horizontal strikes a building 17.0 away at a point 7.00 above the point from which it is thrown. Ignore air resistance.

Find the magnitude of the initial velocity of the baseball (the velocity with which the baseball is thrown).

Find the magnitude of the velocity of the baseball just before it strikes the building.

Find the direction of the velocity of the baseball just before it strikes the building.

pilar22 · May 23, 2008, 11:07 PM

A firefighting crew uses a water cannon that shoots water at 25.0 at a fixed angle of 53.0 above the horizontal. The firefighters want to direct the water at a blaze that is 10.0 above ground level.

How far from the building should they position their cannon? There are two possibilities (); can you get them both? (Hint: Start with a sketch showing the trajectory of the water.)

Curlyben · May 23, 2008, 11:14 PM

Questions merged and moved.

Please refer to this announcement: https://www.askmehelpdesk.com/math-s...-b-u-font.html

Capuchin · May 23, 2008, 11:25 PM

The basic method you should be using here is to split the motion of the object into x and y components. Then applying the equations of motion for x and y motion separately. You can always work out the resultant vector at the end again!

galactus · May 24, 2008, 05:14 PM

I will help with this one and you try the others. If you get stuck, write back. I doubt if you respond back at all anyway, but here goes. You did not specify units, so I will use feet.

Originally Posted by pilar22

A tennis player hits a ball at ground level, giving it an initial velocity of 28.0 at 52.0 above the horizontal.

What are the horizontal and vertical components of the ball's initial velocity?

Horizontal component: $x=28cos(52)t$

Vertical component: $y=28sin(52)t-16t^{2}$

How high above the ground does the ball go?

The max height is achieved when dy/dx=0

$y'=28sin(52)-32t=0$

Solve for t and we get $t=0.69 \;\ seconds$

Plug this into y and get 7.61 feet

How long does it take the ball to reach its maximum height?

As showed before, 69 sec.

What is the ball's velocity at its highest point?

The velocity is 0 at the highest point.

For how long a time is the ball in the air?

Set it equal to 0 and solve the y component for t. You get two values. The good one is 1.38 seconds

When this ball lands on the court, how far is it from the place where it was hit?

Plug the previous answer into the x component and get 23.77 feet

pilar22 · May 24, 2008, 05:47 PM

All these answers were wrong because they aren't in feet. I need big time help! Thanks in advance

Originally Posted by galactus

I will help with this one and you try the others. If you get stuck, write back. I doubt if you respond back at all anyway, but here goes. You did not specify units, so I will use feet.

Horizontal component: $x=28cos(52)t$

Vertical component: $y=28sin(52)t-16t^{2}$

The max height is achieved when dy/dx=0

$y'=28sin(52)-32t=0$

Solve for t and we get $t=0.69 \;\ seconds$

Plug this into y and get 7.61 feet

As showed before, .69 sec.

The velocity is 0 at the highest point.

Set it equal to 0 and solve the y component for t. You get two values. The good one is 1.38 seconds

Plug the previous answer into the x component and get 23.77 feet

galactus · May 24, 2008, 06:00 PM

If it's in metric then just change the gravity constant from 16 to 4.9 and recalculate using the same formulas.

Capuchin · May 24, 2008, 06:02 PM

Originally Posted by pilar22

All these answers were wrong because they aren't in feet. I need big time help! Can you please email me at [email protected]. Thanks in advance

We aren't here to do your work for you. Galactus gave you a sample answer, use it to answer the rest. They're all very similar.

Let us know if you get stuck.

pilar22 · May 24, 2008, 06:07 PM

It still keeps giving me the wrong answer.. can u further help me? PLEASE!!

Originally Posted by galactus

If it's in metric then just change the gravity constant from 16 to 4.9 and recalculate using the same formulas.

Rehaan_genius · Aug 11, 2008, 01:06 AM

All the questions asked by pilar22 are from projectile and circular motion(Kinematics).
I will suggest him to again go through all the theory section of these two topics cause the questions are not so difficult and also trying again and again will help to build up the concept and if still you can't do them then ask!