Ashot-putter throws the shot ( mass = 7.3 kg) with an initial speed of 14.0 m/s at a 35.0° angle to the horizontal. calculate the horizontal distance traveled by the shot if it leaves the athlete's hand at a height of 2.20 m above the ground.

The horizontal distance traveled by the shot is 21.4 m

Explanation:

The position of the shot is given by the position vector r

r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)

Where:

r = position vector at time t

x0 = initial horizontal position

v0 = initial velocity

t = time

α = throwing angle

y0 = initial vertical position

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

Please, see the attached figure for a description of the situation. Notice that when the shot hits the ground, the y-component of the position vector of the shot is -2.20 m (origin of the frame of reference located at the throwing point). Using this data, we can obtain the time at which the shot hits the ground and calculate the x-component of "r final", which is the horizontal distance traveled by the shot. Then:

y = y0 + v0 · t · sin α + 1/2 · g · t²

-2.20 m = 0m + 14.0 m/s · t · sin 35° - 1/2 · 9.8 m/s² · t²

0 = -4.9 m/s² · t² + 14.0 m/s · t · sin 35° + 2.20 m

Solving the quadratic equation:

t = 1.87 s

Now, with this time, we can calculate the x-component of "r final"

x = x0 + v0 · t · cos α

x = 0 m + 14.0 m/s · 1.87 s · cos35°

x = 21.4 m

The horizontal distance traveled by the shot is 21.4 m

f=ma a/f/m a= 480/60 a = 8m/s/s that is quick wow

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