1. Jul 30, 2012

bmguico

1. The problem statement, all variables and given/known data
Problem Number 1: a ball thrown into the air from the ground reaches a maximum height of 50meters and 200meters horizontally. Find the magnitude and direction.

Problem Number 2: a movie stunt drive on a motorcycle speeds horizontally of a 50meter high cliff. How fast must the motorcycle leave the cliff top to land on level ground below 90meters from the base of the cliff where the cameras are? "Ignore air resistance"

2. Relevant equations
X=V(Square Root of 2y/gravity) since there is no time given

3. The attempt at a solution
Please help our professor is asking us to find the magnitude(velocity) and direction(angle) I have no idea how to solve this. For problem number 2 I believe Im looking for velocity so I used the formula above and came up with x=V(square root of 2y/g) or x=V(square root of 2(50m)/9.8m/s square and got the answer 28.21 but I think its wrong because Velocity's unit is meter per second. "Someone please help me" Im really struggling with these 2 questions

2. Jul 30, 2012

azizlwl

It is 2 dimension motion.
Find the equation for horizontal motion and vertical motion.
Remember at time t, the object has moved horizontally and vertically too.

3. Jul 30, 2012

bmguico

4. Jul 30, 2012

voko

The (vector) velocity of the ball or the motorcycle at any monent is the (vector) sum of its (vector) "vertical velocity" and (vector) "horizontal velocity". Is the vertical velocity subject to any acceleration? How about the horizontal velocity? How do the conditions mentioned in the problems on hight/distance translate to conditions on vertical/horizontal velocities?

5. Jul 30, 2012

Delphi51

The trick is that the horizontal and vertical motions are independent. So you can do them separately. Make two headings for horizontal and vertical. Ask yourself what kind of motion takes place each way and write the appropriate equation(s) for each from your formula set. Put in the given numbers; hopefully one of the equations will allow you to find something new.

6. Jul 30, 2012

bmguico

Thank you

7. Jul 30, 2012

bmguico

The problems above are the only given our prof gave us. The formulas that we have are the ff: X=Vcos(theta)T, Y=Vsin(theta)t-1/2gt squared, H=1/2gt squared, T=[2Vsin(theta)]/g, R=[Vsquared sin2(theta)]/g, Y=Vsquaredsinsquared(theta)/2g, Y=1/2gtsquared, X=V square root of 2y/g

Those are all the formulas that I have but I dont have a formula to get the angle for problem number one. If I could just get that I can solve for the magnitude is my answer for number 2 correct? 28.21 is supposed to be the velocity but based on units I think it came out m/s squared and the the correct unit for velocity is meters per second.

8. Jul 30, 2012

voko

 Sorry, I misread the formulas.

Last edited: Jul 30, 2012
9. Jul 30, 2012

voko

You have a formula that relates the max height to Vsin(theta). So you get Vsin(theta). Then you have a formula that relates Vsin(theta) to the time to the top. Note that this time is ONE HALF of the total flight time in the case of the ball going up and then down. Finally, you have a formula that relates Vcos(theta) with the time and distance. With a little trigonometry and algebra, you will get V and theta.

10. Jul 30, 2012

Delphi51

Horizontally you have uniform motion, so the formula is d = V*t or x = Vx*t
Vertically you have accelerated motion due to gravity. Do you have a d = and a v = formula for accelerated motion? I don't like to write them for you because my high school notation may be confusing to you.

If you put the numbers from #1 into x = Vx*t you will immediately be able to find something useful!

11. Jul 30, 2012

azizlwl

For constant acceleration(zero is constant too) you can use this equation for horizontal and vertical motion.

d(displacement)=d0+u0t +0.5at2