Posted by **brenda** on Wednesday, September 22, 2010 at 12:59pm.

A shotputter throws the shot with an initial speed of 15.5 m/s at a 32.0 degree angle to the horizontal. Calculate the horizontal distance traveled by the shot if it leaves the athlete's hand at a height of 2.20 m above the ground.

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**MathMate**, Wednesday, September 22, 2010 at 1:12pm
Split the initial velocity (u) into two components, vertical (uy) and horizontal (ux) using:

ux=u cos(θ) and

uy=u sin(θ)

Solving the time (t) the shot stays in the air from:

Sy=uy*t - (1/2)gt²

where Sy = vertical distance travelled = -2.2m

(reject the negative root).

The horizontal distance travelled is then

Sx=ux*t

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**Sam**, Wednesday, September 22, 2010 at 2:10pm
What do you mean by where Sy = vertical distance traveled = -2.2m? Is the -2.2m from the equation above? Do I add that in to the equation?

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**MathMate**, Wednesday, September 22, 2010 at 5:27pm
-2.2m is the distance between the hand and the ground.

So we equate the vertical distance (from the hand) with the distance as a function of time:

-2.2=uy*t - (1/2)gtē

Solve for t.

- physics -
**self**, Monday, October 19, 2015 at 5:38pm
24.1 m

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