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xxTIMURxx [149]
3 years ago
8

An archer fires an arrow at a castle 230 m away. The arrow is in flight for 6 seconds, then hits the wall of the castle and stic

ks 16 m above the ground. You may ignore air resistance. a) What was the initial speed of the arrow? b) At what angle above horizontal was the arrow shot?
Physics
1 answer:
Vika [28.1K]3 years ago
5 0

Answer:

(a). The initial speed of the arrow is 49.96 m/s.

(b). The angle is 39.90°.

Explanation:

Given that,

Horizontal distance = 230 m

Time t = 6 sec

Vertical distance = 16 m

We need to calculate the horizontal component

Using formula of horizontal component

R =u\cos\theta t

Put the value into the formula

\dfrac{230}{6} = u\cos\theta

u\cos\theta=38.33.....(I)

We need to calculate the height

Using vertical component

H=u\sin\theta t-\dfrac{1}{2}gt^2

Put the value in the equation

16 =u\sin\theta\times6-\dfrac{1}{2}\times9.8\times6^2

u\sin\theta=\dfrac{16+9.8\times18}{6}

u\sin\theta=32.06.....(II)

Dividing equation (II) and (I)

\dfrac{u\sin\theta}{u\cos\theta}=\dfrac{32.06}{38.33}

\tan\theta=0.8364

\theta=\tan^{-1}0.8364

\theta=39.90^{\circ}

(a). We need to calculate the initial speed

Using equation (I)

u\cos\theta\times t=38.33

Put the value into the formula

u =\dfrac{230}{6\times\cos39.90}

u=49.96\ m/s

(b). We have already calculate the angle.

Hence, (a). The initial speed of the arrow is 49.96 m/s.

(b). The angle is 39.90°.

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