Posted by julie on .
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 
bobpursley,
horizontalvelocity=36*cos30
horizDistance=horveloicty*time
time= 41/36cos30
height attained:36sin30*timeabove1/2 g timbabove^2 
physics 
bobpursley,
and adjust that for the starting height 1.4m.

physics 
Damon,
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