Physics 208A. Answers to Test 2 29 April/98

Master

You may use one sheet of helpful information for this test. This information should be made by the student and could contain useful information like conversion factors and formulae as well as example problems. You will be permitted one sheet for each of the tests. No books or notebooks are permitted.

You may leave your answers in unrationalized form. eg Kinetic E = 1/2 * 4 *30^2 Joules. There is no need to use a calculator to multiply out the numbers. The quantities must be in a consistent set of units. The preferred units are the International units….. m, sec, joules and watts. Answers with mixed units will be marked-off.

1. for a vertical leap of 0.8 meter, a 60 kg athlete accelerates herself upwards by bending her knees such that she lowers herself a distance of 0.15 meters, then leaps up.

a) What is the initial velocity at take off?

A person falling from a height of 0.8 meters would have the velocity v^2 = 2gy or v = sqrt(2gy) =sqrt(2*10*0.8) = 4 m/s

Since the jump is just the mirror of this, the initial velocity must also be 4m/s upwards.

b) How long is the jumper in the air? [Hang time].

y = ½*g*t^2. For the fall time, t = sqrt(2*y/g) = sqrt(2*0.8/10) sec = 0.4 seconds

The total time in the air is twice this [rise and fall] = 0.8 sec

Another way of calculating this is to note that the average velocity is 2m/s for the trip [average of 0 and 4] and the distance is 0.8 so the time is t = distance/Vavg = 0.8/2 = 0.4m/s for the trip down . The total round trip is 0.8 sec.

Tennis service.

a) How long does a tennis ball stay within a target region of .05 m if the tossed ball is just near the peak of it's height?

To falls from it’s peak a distance of 0.05m in a time obtained from y = ½*g*t^2. T = sqrt(2y/g) = sqrt(2*.05/10) = 0.1 sec.

Since the ball during the rising portion is within this target range for 0.1 sec, The total hang time in this range is 0.2 seconds.

b) How long does a tennis ball stay within a target region of 0.05m if the tossed ball is tossed high and falls 1.0 meter to reach the target region?

After falling for one meter, the ball has a velocity of v^2 = 2gy . v = sqrt(2*10*1) = 4.5m/s. It falls a distance of approximate 0.05m in a time of t = y/Vavg = .05/4.5 = .01 seconds. This is far, far shorter than the 0.2 seconds of the ball staying in the target window for the case (a). It is much easier to time your racket to strike the ball in case (a) by an overwhelming margin.

3. A long jumper is able to take off with a horizontal velocity of 10 m/s and a vertical velocity of 4 m/s.

a)How far horizontally is the centre of the jumper of the jumper advanced when the cm reaches the same height as at the takeoff point?

Since the takeoff and landing height are the same,

R = 2 VxVy/g = 2*10*4/10 = 8m

How can the jumper make a regulation measured jump that is farther than the physics measured jump described above?

Make a sketch of this.

In this case, the jumper lands with the legs in front so that the jumper is lower, thereby increasing the hang time and the legs make a mark in the sand ahead of the average position of the body.

4.A swimmer wanting to safely cross a rather wide river must decide how to aim himself in his swim. The river is flowing with a

low velocity but the water is smooth and not turbulent at all. The river be 100m wide and the current flows at 1 m/sec. The swimmer can swim [in still water] at 2 m/sec.

What is the shortest time for the journey?

 

If the swimmer aims himself directly transverse to the river’s flow, he will be carried down stream but he will be maximizing his cross river velocity, 2m/s. The time then is t = y/V = 100/2 = 50 seconds

and how far does he go [as measured by an land observer] ?

He will go 100m across the river and during his swim, he will be carried downstream by the current of 50m. x = Vx*t =1*50

5. .If Lewis takes 4 seconds to reach his top speed of 11m/s, what is his average acceleration?

a = Dv/Dt = 11/4 m/s/s = 2.75 m/s/s

6. How fast must a ball be travelling in order for a pitched ball to be almost straight... as defined by the downward deflection is less

than .02m during the 18m flight to the plate?

The fall time must be quite small. Y=1/2 gt^2 t = sqrt(2y/g) = sqrt(2*.02/10) = .063 sec.

Vx = x/t = 18/.063 = 284 m/s!! Close to the speed of sound!!

7. a) What entry angle should a basketball have as it approaches the rim for the smallest possible velocity for a free throw shot?

For any range, the minimum velocity to reach that range is a shot launched at an angle of 45 degrees [equally sharing Vx and Vy] from the same height. The entry velocity must have an angle of 45 degrees.

b)The free throw line is a distance of 5.8 m. Assuming that the ball is released at rim height, what is the minimum launch velocity? of the ball

R = 2Vx Vy/g where Vx and Vy are the same so R = 2Vx^2/g. Vx = sqrt(R*g/2) = sqrt(5.8*10/2) =5.4m/s

The ball must be released with Vx = Vy = 5.4m/s