Posted by Hannah on Monday, May 28, 2012 at 9:25pm.
A golfer, standing on a fairway, hits a shot to a green that is elevated 5.33 m above where she is standing. If the ball leaves her club with a velocity of 48.0 m/s at an angle of 35.7 ° above the ground, find the time that the ball is in the air before it hits the green.
Would I use the equation
y0 + voy t  1/2gt^2??

Physics(Please respond)  Elena, Tuesday, May 29, 2012 at 10:46am
h=5.33 m, v(o) = 48 m/s, α = 35.7º
v(oy) = v(o) •sin α = 48• sin35.7º =28 m/s,
The trajectory of the ball is symmetric with respect to the highest point (let it be the point A). Point B (the point where the ball hits the green), point C is at the rising branch of trajectory, and the initial point O.
Now, time OB (which is the question of the problem) =2•( time OA)  time OC. (To imagine make the scheme of trajectory)
OA:
v(y)= v(oy) –gt.
At the point A
v(y) = 0,
0 = v(oy) –g•t.
time OA
t(OA)=v(o) •sin α/g = 28/9.8 =2.86 s.
time OC =t. The height of the point C is h (as well as the point B)
h = v(oy) •t  gt²/2 = 28•t 4.9•t² = 5.33
4.9•t² 28•t + 5.33= 0
t = {28±sqrt(28²4•4.9•5.33)}/2•4.9 = {28±26}/9.8.
Two roots. We take t= 0.2 because the second root is greater than time OA).
Time OB = 2•2.86 – 0.2 =5.720.2=5.7 s.
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