Posted by **someone** on Sunday, August 2, 2009 at 1:29am.

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.

t=2.06?

x=22.08? am i right?

- Physics -
**drwls**, Sunday, August 2, 2009 at 2:33am
The time the shot rises is Vsin40/g = 0.917s. The distance it rises in that time is t1 = (g/2)(0.917^2) = 4.12m.

To reach the ground, it must fall 6.32 m from the highest point. This requires a length of time

t2 = sqrt(2*6.32m/g) = 1.135s

The total time of flight is t1+t2 = 2.052s

The distance travelled is

V cos40*2.052s = 22.0 m

I used 9.81 m/s^2 for g. We essentially agree.

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