You would like to shoot an orange in a tree and arrow. The orange is hanging 5.0

Question

You would like to shoot an orange in a tree and arrow. The orange is hanging 5.00 m above the ground. On your first try, you fire the arrow at 38.0 m/s at arn angle of 30.0° above the horizontal from a height of 1.30 m while standing 49.0 m away. Treating the arrow as a point projectile and neglecting air resistance, what s the height of the arrow once it has 30.0 5.00 m 1.30 m 49.0 m 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. 121 m 12.6 m 54.8 m 6.8 m 115 m

Explanation / Answer

v0 = 38 m/s at 30 deg ; h = 1.3 m ; d = 49 m

vx = 38 cos30 = 32.91

vy = 38 sin30 = 19 m/s

we know that,

d = vx t

t = d/vx = 49/32.91 = 1.49 s

we know from eqn of motion

S = ut + 1/2 at^2

S = 19 x 1.49 - 0.5 x 9.8 x 1.49^2 = 17.43 m

Height above the ground will be:

H = 17.43 + 1.3 = 18.73 m

Hence, H = 18.73

for the case 2

using the same eqn

5 - 1.3 = 19 t - 4.9t^2

4.9t^2 - 19t + 3.7 = 0

t = 3.67 ; 0.21

D = vx t = 32.91 x 3.81 = 120

D = 32.91 x 0.21 = 6.8 m

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