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# physics problem watch

1. 1. The picture below shows a speedboat towing a swimmer hitched to a parachute. The diagram on the right shows the angles for the relevant forces. The swimmer's mass is 55kg.
(a)What is the tension in the tow rope?
(b)Given that the boat's speed is 12m/s, calculate the additional power of the boat engine to perform this task.

for (a), can i use the difference of the two angle, cuz when add the two vectors up the vertical component should equal to the weight.
which is 55/sin18=1746N. is it right?

2. In a football match player A kicks a high ball from a side line position to the far corner of the field. Player B, at speed v, is to run down the sideline, catch the ball, and score. Players A and B are each a distance l from the goal line, as shown below, and B is running the instant A kicks.
Show that, if the player's heights and the air resistance of the ball are ignored, the ball must reach height h, and no higher. where h is given by
h=l²*g/(8v²)

here's my working:
time for the ball to reach sideline = time for B to run to the sideline.
t=l/v (for B to run down to sideline)
t=h/(0.5(u+v)) distance over average speed
u=v+at, v=u-at ball goes way up and way down
so they have different u and v at way up and down
substitue u=v+at and v=u-at into t=h(0.5(u+v)) individually
it would come out like t=2h/(2u-at), t=2h/(2v+at) on way up and way down.
u=o, v=0 at up and down.
therefore, t=2h/at and t=2h/at
add them up, the total time for the ball to go up then down.
t=4h/at, substitue t=l/v into it. t=4hv/gl where a=g
finally....... 4hv/gl=l/v, h=gl^2/4v^2
i think i am on the right track, but don't know why i got 4 instead of 8......

could some1 help me~
Attached Images

2. I try the first question but getting the different answer:
Firstly, I draw the force-triangle for person because this person is not accelerated at all.
Using the Sine's rule: a/sinA = b/sinB = c/sinC
Find out the value of angle in triangle: 26, 41, 113
So the tension is: T = mg*sin41/sin26
It's 807 N
3. let the upthrust force on the parachut be U. Now resolving vertically and horizontally:

mg+Tcos67=Ucos41 (vertically)...(1)
Tsin67=Usin41 (horizontally)->U=1.403T...(2)

sub (2) into (1) and solve for T.

539+0.391T=1.06T
0.668T=539
T=807N to 3SF
4. (Original post by VCVT17)
I try the first question but getting the different answer:
Firstly, I draw the force-triangle for person because this person is not accelerated at all.
Using the Sine's rule: a/sinA = b/sinB = c/sinC
Find out the value of angle in triangle: 26, 41, 113
So the tension is: T = mg*sin41/sin26
It's 807 N
very nice method
5. thanks, that's really helpful.
but what about no.2? is my working correct?
6. v=u+at (1) and v^2=u^2 +2as (2)

time = l/v for ball to be caught, therfore t=l/2v, for ball to reach max height.
using (1), 0=u-gl/2v, where u is the initial velocity of the ball.

However, you require an eqtn that does not involve u, therefore need to eliminate it...using (2), 0=(gl/2v)^2 -2gh (remembering directions upwards positive and gravity downwards, therfore its -g)

thus on rearranging: h=gl^2/8v^2, as required.

took me a while but got it in the end, lol,

pk

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