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

Question

You would like to shoot an orange in a tree with vour 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 30.0% above the horizontal from a height of 1.30 m while standing 53.0 nm away. Treating the arrow as a point projectile and neglecting air resistance, what is the height of the arrow once it has traveled the 53.0 m horizontally to the tree? 5.00 m 1.30 m 53.0 m Number Im 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 54.4 m 89.2 m 16.9 m 18.3 m 87.8 m

Explanation / Answer

part 1:

initial height=1.3 m

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

vertical acceleration=-9.8 m/s^2

initial horizontal speed=33*cos(30)=28.58 m/s

horizontal acceleration=0

let time taken to horizontal distance covered 53 m be t seconds.

then 53=initial speed*time+0.5*acceleration*time^2

==>53=28.58*t

==>t=1.8545 seconds

height of the arrow at this height=initial vertical speed*time+0.5*acceleration*time^2

=16.5*t+0.5*(-9.8)*t^2

=13.747 m

part 2:

let at time t , vertical height achieved be 5-1.3=3.7 m

then 3.7=16.5*t+0.5*(-9.8)*t^2

solving for t , we get

t=3.12577 seconds

then distance from the tree required=28.58*t

=89.3356 m

so second option is correct.

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