physics

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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 36.0 m/s at an angle of 30.0° above the horizontal from a height of 1.40 m while standing 41.0 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 41.0 m horizontally to the tree?

• physics -

horizontalvelocity=36*cos30

horizDistance=horveloicty*time
time= 41/36cos30

height attained:36sin30*timeabove-1/2 g timbabove^2

• physics -

and adjust that for the starting height 1.4m.

• physics -

u = 36 cos 30 = 31.18 m/s forever
d = u t = 41
so t = 41/31.18 = 1.315 seconds in air

Vi = 36 sin 30 = 18 m/s

h = Hi + Vi t - 4.9 t^2
h = 1.4 + 18(1.315) - 4.9 (1.315)^2
= 16.6 meters high

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