Exam Practice Problems: Solving for Time, Speed, and Velocity

[burg]

These are questions from my previous tests within this semester that I didn't exactly get correct. I can't seem to figure them out/find an example, nor is my teacher going to be available before my exam. So here I am. I need to know how to solve them, and them to be solved. These are not homework or assignment problems, these are just very plausible problems for the exam that I want to understand.

1) A football quarterback attempts a pass to one of the receivers. As the ball is snapped, the reciever leaves the line of scrimmage and runs directly down field. The quarterback releases the ball 2.0s later and from a position 3.0m behind the line of scrimmage. He throws the ball with a speed of 26m/s at an elevation of 60degrees above the horizontal. The reciever makes the diving reception, catching the ball just as it reaches the ground. The quarterback is 2.0m tall *where the ball is released*
a) What is the time of flight of the football?
b) What is the average speed of the receiver?

2) A rock of mass 4.0 * 10^2g is tied to one end of a string that is 2.0m in length. Holding the other end above his head, a boy swings the rock around in a circle whose plane is parallel to the ground.
a) If the string can withstand a maximum tension of 4.5N before breaking, what angle to the vertical does the string reach just before breaking?
b) At what speed is the rock traveling just as the string breaks?

3) A helicopter is rising vertically at a uniform velocity of 14.7m/s. When it is 196m from the ground, a ball is projected from it with a horizontal velocity of 8.5m/s with respect to the helicopter. Calculate the following.
a) When the ball will reach the ground?
b) Where it will hit the ground?
c) What its velocity will be when it hits the ground?

Again, please put how it was done. Any help is appreciated. Thanks a lot.

 

[Curious3141]

Why don't you post your own idea of how you would approach them first ? We'd be a lot more amenable to assisting you then.

 

[burg]

Well, for some I don't have any idea really, but here goes..

1a) d = V1(t) + 1/2a(t)^2 This should give me the total time the ball was in the air, I think. But will it give me the total time? Because the ball is thrown 2m above the ground, but caught at the ground.

1b) Find the total distance traveled by the receiver than divide that by the time the ball was in the air + 2.0s. Not to sure which formula.

2) I got 0/6 on my test for this one...but I think it has something to do with the force of gravity down, and the force of tension at an angle. When the combined gravity and tension reach 4.5N, that is the angle where it breaks. How to solve for that I have no idea.
b) Easy once I can do a...

3a) d = V1(t) + 1/2a(t)^2 The helicopter is moving up, therefore it will have the V1 of the helicopter, the a of gravity, and d 196. No?
b) Dh = V1h(t) + 1/2a(t)^2 Using the horizontal component. Should give the you the distance away from the helicopter, No?
c) V2^2 = V1^2 + 2ad Using the y components, so give the velocity, No?

I don't know the answers, so I have no idea if I'm on the right path or not. Thanks.

 

[Curious3141]

Well, for some I don't have any idea really, but here goes..

1a) d = V1(t) + 1/2a(t)^2 This should give me the total time the ball was in the air, I think. But will it give me the total time? Because the ball is thrown 2m above the ground, but caught at the ground.

I have a small nit to pick with the question. It was never specified whether the thrower threw the ball directly parallel to the line of scrimmage or at an angle to it. I'm assuming it's a direct sideways pass parallel to the scrimmage line.

As you said the ball starts out 2 meters above the ground. Hence the formula you use should have an initial term to correct for this. Try :

y_t = y_0 + (v\sin\theta)t - \frac{1}{2}gt^2 and figure out what each term should be, then solve for the time.

1b) Find the total distance traveled by the receiver than divide that by the time the ball was in the air + 2.0s. Not to sure which formula.

Try total distance/total time (you're on the right track).

2) I got 0/6 on my test for this one...but I think it has something to do with the force of gravity down, and the force of tension at an angle. When the combined gravity and tension reach 4.5N, that is the angle where it breaks. How to solve for that I have no idea.
b) Easy once I can do a...

Consider the forces acting on the mass. There are two, the tension in the string and gravity pulling vertically downwards. Resolve the tension into a horizontal and vertical component. Which component opposes the weight of the mass ? Can you form an equation to solve for the angle ?

Once you get that angle from the first part, use the horizontal component of the tension to equate to the centripetal force on the mass (F = \frac{mv^2}{r}). Here, don't forget that the radius of revolution of the mass is not the length of the string, but less. You can form a trigonometric expression to relate the radius of revolution to the length of the string, and proceed to solve for the velocity of the mass when the string just breaks.

3a) d = V1(t) + 1/2a(t)^2 The helicopter is moving up, therefore it will have the V1 of the helicopter, the a of gravity, and d 196. No?
b) Dh = V1h(t) + 1/2a(t)^2 Using the horizontal component. Should give the you the distance away from the helicopter, No?
c) V2^2 = V1^2 + 2ad Using the y components, so give the velocity, No?

I don't know the answers, so I have no idea if I'm on the right path or not. Thanks.

Here you're given the particle's initial horizontal and vertical velocities, along with the initial vertical displacement. You're forgetting to add the initial vertical displacement of the ball in your vertical equation (otherwise it looks right). Use the same method I specified in the first problem to figure it out.

For the horizontal motion, if you neglect air resistance, the acceleration should be zero.

You should be able to figure out how to do the first two parts.

For the third part I would suggest using v = u + at to get the final vertical component of the velocity of the ball just before it hits the ground. The horizontal component of the velocity is what it was initially (no air resistance). For the answer you need to do a vector addition of the two quantities. For the magnitude (speed) you can use Pythagoras' theorem, but you need to specify a direction too. Can you think of an expression to use for that ? (Hint : think of the tangent of the angle of trajectory and how it relates to the vertical and horizontal components of velocity).

Post again if you need further help, but only after you've tried these out and really thought about it. Good luck.

 

[burg]

Exam was about an hour ago, quite easy actually. The info for one you told me proved useful :) Thanks

 

[Curious3141]

I'm glad. Hope you do well.