Question

You would like to shoot an orange in a tree with your bow and arrow. The orange is hanging 5.00 m above the ground. On your first try, you fire the arrow at 32.0 m/s at an angle of 30.0° above the horizontal from a height of 1.30 m while standing 54.0 m away. Treating the arrow as a point projectile and neglecting air resistance, what is the height of the arrow once it has traveled the 54.0 m horizontally to the tree?If you fire at the same speed and angle on your second try, how far away could you stand such that the arrow will hit the orange? Assume that the orange remains fixed in place during the arrow\'s flight. Select all that apply.

Answer 1

a)

32 m/s @ 30° gives

Vx = 27.71 m/s

Vy = 16 m/s

now find time

x = Vx t

54 = 27.71 t

t = 1.95 s ( tome to get to the tree)

then the height

d = Vi t + 1/2 g t^2

d = 16*(1.95) + 1/2 (-9.81) (1.95^2) = 12.55 m (height above release point)

height above ground = 12.55 + 1.3 = 13.85 m

b)
now we solve for time to height of 1.3 m

d = Vi t + 1/2 g t^2

1.3 =16t - 4.905 t^2

4.905 t^2 - 16 t + 1.3 = 0

t = 3.18 s or 0.0833 s

at the shorter time the arrow is on the way up and at the longer it is coming down

so x = Vx t = 27.71(3.18) = 88.11 m

or x = 27.71 (0.0833) = 2.3 m