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

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 33.0 m/s at an angle of 30.0° above the horizontal from a height of 1.2 m while standing 57 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 57.0 m horizontally to the tree? Sapling Learning Map 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 33.0 m/s at an angle of 30.0° above the horizontal from a height of 1.20 m while standing 57.0 m away. Treating the arrow as a point 30.0° 5.00 m jectile and neglecting air resistance, what s the height of the arrow once it has traveled the 57.0 m horizontally to the tree? 1.20 m Number 57.0m If you fire at the same speed and angle on your second try, how far away could you stan that the arrow will hit the orange? Assume that the orange remains fixed in place d such arrow's flight. Select all that apply 7.1 m 9.6 m 86.5 m 86.5m 59.5 m 89.0 m Give Up & View Solution 0 Check Answer 0 Next -el Exit -- O Previous Hint

Explanation / Answer

part a:

initial horizontal speed=33*cos(30)=28.5788 m/s

time taken to cover the horizontal distance of 57 m=57/28.5788=1.99448 seconds

initial vertical speed=33*sin(30)=16.5 m/s

assuming vertically upward direction to be positive, acceleration due to gravity=-9.8 m/s^2

vertical distance travelled in 1.99448 seconds=initial speed*time+0.5*acceleration*time^2

=16.5*1.99448-0.5*9.8*1.99448^2=13.417 m

then height from the ground=1.2+13.417=14.617 m

part b:vertical distance to be covered=5-1.2=3.8 m

initial vertical speed=16.5 m/s

let time taken to travel 3.8 m is t seconds.

then 16.5*t-0.5*9.8*t^2=3.8

==>4.9*t^2-16.5*t+3.8=0

solving for t , we get t=0.2486 seconds or t=3.11868 seconds

if we take t=0.2486 seconds, horizontal distance=33*cos(30)*t=7.1047 m

if we take t=3.11868 seconds , then horizontal distance=33*cos(30)*t=89.1282 m

so both first option (7.1 m) and last option (89 m) are applicable.

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