Question

An arrow is shot from a bow at 20.0 m/s at an angle of 65º above the horizontal. The arrow leaves the bow at a height of 1.80 m. At what height will it strike a wall that is 10.0 m away? (Remember to connect the two motions of the arrow using time.)

y = _____ m

Answer 1

we need to find the components of the vector.

using trigonometry:

\(\sf sin\theta=\dfrac{opposite}{hypotenuse}\) \(\sf cos\theta=\dfrac{adjacent}{hypotenuse}\) \(\sf tan\theta=\dfrac{opposite}{adjacent}\)

the velocity in the y direction is: \(\sf sin\theta=\dfrac{v_y}{20~m/s}\rightarrow v_y=sin\theta(20~m/s)\)

the velocity in the x direction is \(\sf cos\theta=\dfrac{v_x}{20~m/s}\rightarrow v_x=cos\theta(20~m/s)\)

Because these are independent of each other, e can find how long it takes for the arrow to reach the wall (10m) and use it to find the height at that instant.

Horizontal trajectory:

\(\Delta x=v_x\Delta t+\dfrac{1}{2}a(\Delta t)^2\)

a= 0, \(\Delta x\)=10 m

\(10~m=cos\theta(20~m/s)\Delta t\rightarrow \Delta t=\dfrac{10~m}{cos\theta(20~m/s)}\)

Now that we know the time, we can find the height when it hits the wall, \(y_f\).

Vertical trajectory:

\(y_f=y_i+v_y\Delta t+\dfrac{1}{2}a(\Delta t)^2\)

a=-9.8 m/s, \(y_i\)=1.8 m, \(\Delta t=\dfrac{10~m}{cos\theta(20~m/s)}\)

\(y_f=1.8~m+sin\theta(20~m/s)(\dfrac{10~m}{cos\theta(20~m/s)})-\dfrac{9.8~m/s^2}{2}(\dfrac{10~m}{cos\theta(20~m/s)})^2\)