A shotputter throws the shot with an initial speed of 16.0m/s at a 45.0 angle to the horizontal. Calculate the horizontal distance traveled by the shot if it leaves the athlete's hand at a height of 2.05m above the ground.

Question

agent0smith · Answer

I'd start by breaking down the velocity into horizontal and vertical components, then finding the time it's in the air from vertical velocity, and then once you have t, you can multiply it by the horizontal velocity to find the horizontal distance.

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anonymous · Answer

wow. thank you so much
The answer I got was 28.0327445, which was correct

agent0smith · Answer

To find t, use the fact that it'll hit the ground 2.05m below the starting point, so the vertical displacement y = -2.05
$$y=v _{y}t-\frac{ 1 }{ 2 }g t ^{2}$$
Vy is the vertical velocity from the above drawing. You'll get a quadratic equation which you need to solve for t.

Once you have t, then it's easy to find the horizontal displacement x$$x=v _{x}t$$

agent0smith · Answer

Oh you already got it haha! Good :)

anonymous · Answer

I just used exactly what you said. Congrats sir, you just earned yourself a fan.

agent0smith · Answer

haha, I was still typing in the method while you already worked it out. Nice job!

agent0smith · Answer

I checked your answer just to be sure, and you were correct :)