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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?

Answer 1

Emily Abbate

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

(b). #21.72"m"#

(a).

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

#v=u+at#

This becomes:

#0=u+"g"t#

#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=vcos(60)#

#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 #=vcos(30)#

So the distance #=vcos(30)xxt#

#=15.68xx0.866xx1.6=21.72"m"#

To get the height reached:

#v^2=u^2+2ah#

#0=7.84^2-2xx9.8xxh#

#h=3.16"m"#

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

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

#21.72/2=10.86"m"# from the archer.

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Answer 2

Adrian Ayala

The initial speed of the arrow was approximately 12.8 m/s. The target was approximately 14.7 meters away from the archer.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.