Posted by **Kate** on Tuesday, August 9, 2011 at 2:57pm.

A discus is thrown from a height of 3 feet with an initial velocity of 55 ft/s at an angle of 44° with the horizontal. How long will it take for the discus to reach the ground?

I started off with 55cos(44)=40

then I made an equation

40t-16(t)=2.5 seconds?

I feel like I did this wrong because I didn't add 3 feet anywhere

or do I add 3 in to 55cos(44) giving me an answer of 2.6

- Precalculus -
**MathMate**, Tuesday, August 9, 2011 at 5:09pm
Yes, the 3 feet are important.

You can consider the datum (y=0) at 3' from the ground, and the ground is therefore at -3 feet.

The vertical initial velocity, vy, is u*sin(θ), so it gives

vy=55sin(44°)=38.21 approx.

The (vertical) distance travelled is given by:

S=vy*t-(1/2)gt²

where

S=-3 (ground)

vy=55sin(44°), and

g=32.2 ft/s²

Solve for t.

I get -0.08 and 2.45 s.

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