Question

You are watching an archery tournament when you start wondering how fast an arrow is shot from the bow. Remembering your physics, you ask one of the archers to shoot an arrow parallel to the ground. You find the arrow stuck in the ground 61.0 m away, making a 2.00 ∘ angle with the ground?

Answer 1

Answer:

Velocity of throwing arrow = 43.13 m/s.

Explanation:

In the question,

Let us say the height from which the arrow was shot = h

Distance traveled by the arrow in horizontal = 61 m

Angle made by the arrow with the ground = 2°

So,

From the equations of the motion,

$61 =u.t\\t=\frac{61}{u}$

Now,

Also,

Finally, the angle made is 2 degrees with the horizontal.

So,

Final horizontal velocity = v.cos20°

Final vertical velocity = v.sin20°

Now,

u = v.cos20° (No acceleration in horizontal)

Also,

$v=u+at\\vsin20=0+9.8(t)\\t=\frac{v.sin20}{9.8}$

So,

We can say that,

$\frac{v.sin20}{9.8}=\frac{61}{v.cos20}\\v^{2}.sin20.cos20=597.8\\v^{2}=1860.56\\v=43.13\,m/s$

Therefore, the velocity with which the arrow was shot by the archer is 43.13 m/s.

Answer 2

Velocity of a object is the ratio of distance traveled by the object with the time taken. The speed of the arrow shot is 130.94 m/s.

What is velocity of a object?

Velocity of a object is the ratio of distance traveled by the object with the time taken.

Given information-

The arrow stuck in the ground 61.0 m away.

The angle made by the arrow with the ground is 2 degrees.

Convert this angle into the radians as,

$\theta=\dfrac{\pi}{180}\times2 \\\theta=0.0349 \rm rad$

Thus, the angle made by the arrow with the ground is 0.0349 radians.

The ratio of the vertical velocity $v_y$horizontal velocity $v_x$ is,

$tan\theta=\dfrac{v_y}{v_x}=\theta \rm (for \;smaller \;angle)$

Put the values as,

$\dfrac{v_y}{v_x}=0.0349\\v_y=0.0349v_x$          .....1

Let the above equation as equation 1.

Time taken by the arrow to traveled the distance of 60 meters with horizontal velocity is,

$t=\dfrac{61}{v_x}$

Now the vertical velocity can be given as,

$v_y=gt$

Put the value of vertical velocity from equation 1 and the value of time in the above equation as,

$0.0349v_x=9.81\times\dfrac{61}{v_x}\\v_x=130.94\rm m/s$

Hence, the speed of the arrow shot is 130.94 m/s.

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