Question

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 sticks 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?

Answer 1

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

(b). The angle is 39.90°.

Step-by-step 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

`(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-(1)/(2)gt^2`

Put the value in the equation

`16 =u\sin\theta*6-(1)/(2)*9.8*6^2`

`u\sin\theta=(16+9.8*18)/(6)`

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

Dividing equation (II) and (I)

`(u\sin\theta)/(u\cos\theta)=(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* t=38.33`

Put the value into the formula

`u =(230)/(6*\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°.