You would like to shoot an orange out of a tree with your bow and arrow. The ora

Question

You would like to shoot an orange out of 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 33.0 m/s at an angle of 30.0 degree above the horizontal from a height of 1.10 m while standing 52.0 m away. Treating the arrow as a point projectile and neglecting air resistance, what is the height of the arrow once it has travelled the 52.0 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? Select all that apply

Explanation / Answer

(a)
Height of Orange = 5.0 m
Speed = 33 m/s

Horizontaly Velcoity = 33 * cos(30) = 28.57 m/s
Vertical Velcoity = 33 * sin(30) = 16.5 m/s

Time taken by Arrow to travel 55 m horizontal distance =
t = Distance/speed
t = 55/ 28.57
t = 1.92s

Vertical Distance covered in this time ,
S = u*t - 0.5at^2
S = 16.5 * 1.92 - 0.5*9.8*1.92^2
s = 13.6 m

Height of arrow = 13.6 + 1.10 = 14.7m

(b)
Initial Velocity and angle is kept same.

Horizontaly Velcoity = 33 * cos(30) = 28.57 m/s
Vertical Velcoity = 33 * sin(30) = 16.5 m/s

Let you stand x distance away from tree.
Time taken by Arrow to travel distance x, t = x/28.57 s

This time should be equal to the vertical distance covered by arrow = 5.5-1.1m = 3.9 m

S = u*t - 0.5*a*t^2
3.9 = 16.5*(x/28.57) - 0.5*9.8*(x/28.57)^2

Solving For X -

x = 7.3m & 88.8m

Horizontal Distance at which arrow will hit the orange , x = 7.3m & 88.8m

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