Posted by someone on .
a shotputter 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,
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.