An archer shoots an arrow at an angle of 30 degrees from a height of 1.4m. If it took 1.6 seconds for the arrow to hit the target, which is 1.4m tall, what was the arrow's initial speed? How far away was the target?

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Answer 1 · 2015-09-21T12:48:47

(a). $15.68 m/s$ $15.68"m/s"$

(b). $21.72 m$ $21.72"m"$

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(a).

To get the initial vertical component of velocity we can use:

$v = u + a t$ $v=u+at$

This becomes:

$0 = u + g t$ $0=u+"g"t$

$0 = u - 9.8 \times 0.8$ $0=u-9.8xx0.8$

(0.8 = 1.6/2 - which is the time taken to reach the zenith)

$u = 7.84 m/s$ $u=7.84"m/s"$

$u = v cos (60)$ $u=vcos(60)$

$v = \frac{u}{cos (60)} = \frac{7.84}{0.5} = 15.68 m/s$ $v=u/cos(60)=7.84/0.5=15.68"m/s"$

(b).

The distance covered = speed x time

The horizontal component of the velocity $= v cos (30)$ $=vcos(30)$

So the distance $= v cos (30) \times t$ $=vcos(30)xxt$

$= 15.68 \times 0.866 \times 1.6 = 21.72 m$ $=15.68xx0.866xx1.6=21.72"m"$

To get the height reached:

$v^{2} = u^{2} + 2 a h$ $v^2=u^2+2ah$

$0 = {7.84}^{2} - 2 \times 9.8 \times h$ $0=7.84^2-2xx9.8xxh$

$h = 3.16 m$ $h=3.16"m"$

So the total height above the ground $= 3.16 + 1.4 = 4.56 m$ $=3.16+1.4=4.56"m"$

It reaches maximum height 1/2 way through the flight so it will be

$\frac{21.72}{2} = 10.86 m$ $21.72/2=10.86"m"$ from the archer.