Question

An athlete executing a long jump leaves the ground at 28.0 angle and travels 7.80 m
a)What was the take off speed?
b)If this speed were increased by just 5.0%,how much longer would the jump be?

Answer 1

Ok.. so.. we are working in two dimensions here

so can u write down the given data.. (like ur teacher said).. here?

Answer 2

ok

Answer 3

To find: final velocity vo=?

timet)=?

Given:Angle of projection=28.0deg

initial velocity=0m/s2

acceleration:9.8m/s^2

displacement y-yo=7.80m

Answer 4

ps. initial velocity is not zero.. else how woudl he jump? :P

Answer 5

but didn't he start at 0 before he jumped?

Answer 6

but.. u are applying the equation the moment.. THE MOMENT he jumps right?
so in that case, the initial velocity is the velocity of junp

just like when he lands.. he doesn't land with zero speed right?.. he lands with some speed.. and THEN the speed goes to zero.. got what i mean?

Answer 7

yes

Answer 8

so with this we are only dealing with what happends when he actually moved?

Answer 9

when he is in air!..

Answer 10

oh ok

Answer 11

my acceleration would be positive?

Answer 12

we are in two dimensions here..

so you need to resolve and work in each direction x and y seperately

Answer 13

ok.i understand. so in this I am finding the time and initial velocity right?

Answer 14

yup..

Answer 15

ok cool:) so in the x axis what is given is the angle of projection which is 28.0 deg and acceleration which is 9.8m/s^2 and in our y axis we are given the displacement y-yo=7.80m

Answer 16

*just to let you know I really appreciate u helping me understand this .I did really bad on the first test and need to pass the other 2