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.

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

okay so t=2.06?

so x=22.08?

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.

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..

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

so which should i use 4.4 or 5.34?

which is slighty over 1.8seconds anyway? and shud i just subtract it?

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

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

Isn't there a negative sign before the 4.9?

To calculate the horizontal distance traveled by the shot, we can use the following equations:

1. The vertical component of velocity (Vy) at any time t is given by:
Vy = V0 * sin(θ), where V0 is the initial speed and θ is the launch angle.

2. The time taken to reach maximum height (t_max) can be found using the equation:
t_max = Vy / g, where g is the acceleration due to gravity (approximately 9.8 m/s^2).

3. The maximum height (H_max) reached by the shot is:
H_max = Vy^2 / (2 * g).

4. The total time of flight (t_total) can be calculated using:
t_total = 2 * t_max.

5. Finally, the horizontal distance traveled (R) is given by:
R = V0 * cos(θ) * t_total.

Now, let's calculate the horizontal distance traveled by the shot:

Given:
V0 = 14 m/s
θ = 40 degrees
H = 2.2 m

1. Calculate Vy:
Vy = 14 * sin(40) = 9.001 m/s.

2. Calculate t_max:
t_max = 9.001 / 9.8 = 0.918 s.

3. Calculate H_max:
H_max = (9.001^2) / (2 * 9.8) = 4.167 m.

4. Calculate t_total:
t_total = 2 * 0.918 = 1.836 s.

5. Calculate R:
R = 14 * cos(40) * 1.836 = 47.17 m (rounded to two decimal places).

Therefore, the horizontal distance traveled by the shot is approximately 47.17 meters.