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 36.0 m/s at an angle of 30.0° above the horizontal from a height of 1.40 m while standing 56.0 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 56.0 m horizontally to the tree? 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.

Explanation / Answer

Hi,

In this case we have to remember a little about parabolic movement. As they are telling us that we can ignore the air resistance and we can consider the arrow as a point projectile then we can use the following set of equations:

Vox = Vo cos (1) ; Vx = Vox (2) ; x =  Vox t (3)

Voy = Vo sin (4) ; Vy = Voy - g t (5) ; y = yo + Voy t - (1/2) g t2 (6)

Where Vo is the initial velocity, is the angle above the horizontal, t is the time, y is the vertical distance, x is the horizontal distance and g is acceleration of gravity which will have a value of 9.8 m/s2 in this problem.

For starters, we use the equations (1) and (4) to find the components of the initial velocity in each axis:

Vox = (36 m/s)*cos(30) = 31.2 m/s

Voy = (36 m/s)*sin(30) = 15 m/s

In the first part they are asking us the value of the vertical distance (y) when the horizontal distance (x) is equal to 56 m, so we use the equation (3) to find the time and then the equation (6) to find the height.

t = x/Vox = (56 m)/(31.2 m/s) = 1.8 s

y = (1.4 m) + (15 m/s)*(1.8 s) - (1/2)*(9.8 m/s2)*(1.8 s) = 12.5 m

For the second part they are asking us the horizontal distance (x) if we want to shoot the orange (which has a height of 5 m) using the same initial velocity and the same angle.

In order to do that we use the equation (6) to find the time. This will give us two values. If one of them is negative then we only use the other one. If both of them are positive then we must use both of them because one corresponds to the first half of the parabolic movement while the other is related to the second one. The equation (3) is used to find the distance between us and the orange.

(1/2)*g*t2 - Voy t + (y - yo) = 0 ::::::: 0.5*9.8*t2 - 15*t + (5 - 1.4) = 0 [s] :::::::: 4.9 t2 - 15 t + 3.6 = 0

Solving the previous equation we have the following: t = 0.263 s and t = 2.8 s

The two possible distances are then: x = 8.2 m and x = 87.4 m respectively

I hope it helps

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