Question

An athlete performing a long jump leaves the ground at a 29.6∘ angle and lands 7.68m away.
Part A
What was the takeoff speed?
Express your answer using three significant figures and include the appropriate units.
v =

Answer 1

Part B
If this speed were increased by just 3.0%, how much longer would the jump be?
Express your answer using two significant figures and include the appropriate units.
Δd =

Answer 2

do we use that same formula here too

Answer 3

Same thing as before:\[d=\frac{ v_0^2 }{ g }\sin(2\theta)\]You have the distance, you have the angle and you know g. the rest is easy

Answer 4

7.69 = (sin (2 x 29.6) X 9.8 ) / 2

Answer 5

wrong, review the formula

Answer 6

sorry 7.69 = v0 (sin (2 x 29.6) / 9.8

Answer 7

Also wrong, it is squared velocity

Answer 8

The formula to get velocity is \[v_0=\sqrt{\frac{ d·g }{ \sin(2\theta)}}\]

Answer 9

ohh whoops, 7.69 x 9.8 / sin (2 x 29.6) all squared root

Answer 10

\[v_0=\sqrt{\frac{ 7.68·9.81 }{ \sin(2·29.6)}}\]

Answer 11

i got an answer of 12.352

Answer 12

For part number 2: Now you have a new speed that is a 3% faster than the initial speed\[v_1=1.03v_0\rightarrow d_1=\frac{ v_1^2 }{ g}\sin(2\theta)=1.03^2\left( \frac{ v_0^2 }{g} \sin(2\theta) \right)=1.03^2·d\]

Answer 13

\[\Delta d=\frac{ d_1-d }{ d }=\frac{ (1.03^2-1)d }{ d }=1.03^2-1=0.06090\]And that means distance has increased in a 6.09%

Answer 14

you got the 1.03 bc its 100% plus 3%, but how do you know to add it with 100%

Answer 15

If the new velocity is 103% the initial velocity then you have to multiply by 103 and divide by 100, in other words, multiply by 1.03

Answer 16

ohh what formula would i use for part b then? the same range formula?

Answer 17

Please, read my explanation for part 2 above

Answer 18

ohh okay i got the 3% part but i still dont understand the 2nd part, is this what you are trying to do... d= 1.03 sin(2 x 29.6) / 9.8 = 0.09

Answer 19

Same formula and new speed. now speed is V1=1.03·V0 and that new speed will give a new distance d1

Answer 20

so the speed we found in part a we use for this part b to get d=?

Answer 21

but where do i use the 1.03 since that is a percentage

Answer 22

Do you need to use it? dont you see that it is not necessary? Please, read what I have done carefully

Answer 23

Ther are two posts where I describe in detail how to calculate what the problem is asking for in part 2. Please, come back to me when you read it carefully. You need to do some effort to try to understand what is written

Answer 24

ohh so after finding d we multiply it by the 1.03 bc it increased in 3% ..right?

Answer 25

Well let us leave it for today. You still do not want to read carefully what is written. Nobody has multiplied the distance by 1.03

Have a good night!

Answer 26

i am looking at the formula and trying to understand it but its not sinking in

Answer 27

@ganeshie8 do have another way of explaining part b, i am trying to figure this out and carlosGP helped me out but i still dont understand part b