a shot-putter throws the shot with an intial speed of 14 m/s at a 40 degree angle to the horizontal. Calculate the horizontal distance traveled by the shot if it leaves the athlete's hand at a height of 2.2m above the ground.

Question

i got to answers which is corret?
47.17m or 57.24m or neither?

Answer 1

What is the vertical velocity component?
how long will it stay in the air?

hfinal=hinitial+Vv*timeinair-1/2 g timinair^2

solve for time in air.

Then horizontal distance=Vh*timeinair

I didn't get your answers.

Answer 2

What is the vertical velocity component?
that's 9 m/s right?

the time i got is 4.4/5.34s
which i think are wrong..

Answer 3

correct. I didn't check the quadreatic solutions, but it is slighly over 1.8 seconds

Answer 4

so which should i use 4.4 or 5.34?
which is slighty over 1.8seconds anyway? and shud i just subtract it?

Answer 5

Neither of your answers is correct. The shot returns to 2.2 m height after travelling
X = (V^2/g)sin(2*40) = 19.7 m horizontally. It will only travel another few meters before hitting the ground, about 1/4 second later. BobP's quadratic solution will give the exact value

Answer 6

so the quadratic solution is like this 4.9t^2+9t+2.2=0?

Answer 7

Isn't there a negative sign before the 4.9?

Answer 8

okay so t=2.06?

so x=22.08?