Answer:
a) 21.612 m
b) 34.489 m
c) 15.887 m/s
Explanation:
This situation is related to projectile motion and the main equations that will be useful in this case are:
[tex]y=y_{o}+V_{o}sin\theta t-\frac{gt^{2}}{2}[/tex] (1)
[tex]V^{2}={V_{o}}^{2} - 2gy[/tex] (2)
Where:
[tex]y[/tex] is the height of the arrow at a given time
[tex]y_{o}=1.7m[/tex] is the initial height of the arrow
[tex]V_{o}=26 m/s[/tex] is the initial speed of the arrow
[tex]\theta=60\°[/tex] is the angle at which the arrow was shot
[tex]t=3.4 s[/tex] is the time at which the arrow lands on the top edge of the cliff
[tex]g=9.8m/s^{2}[/tex] is the acceleration due gravity
[tex]V[/tex] is the final speed of the arrow at a given time
Knowing this, let's begin with the answers:
We are told the arrow lands on the top edge of the cliff with height [tex]H[/tex] at [tex]t=3.4 s[/tex]. This means we have to find [tex]y[/tex] at this given time using (1):
[tex]y=H=1.7 m+(26 m/s)sin(60\°)(3.4 s) - \frac{(9.8m/s^{2})(3.4 s)^{2}}{2}[/tex] (3)
[tex]H=21.612 m[/tex] (4) This is the height of the cliff
The arrow has its maximum height when [tex]V=0[/tex], just at the top of its parabolic movement, before falling down. In this case, we will use equation (2):
[tex]0={V_{o}}^{2} - 2gy[/tex] (5)
Isolating [tex]y[/tex]:
[tex]y=\frac{{V_{o}}^{2}}{2g}[/tex] (6)
[tex]y=\frac{(26 m/s)^{2}}{2(9.8 m/s^{2})}[/tex] (7)
[tex]y_{max}=34.489 m[/tex] (8) This is the maximum height of the arrow
If we want to know the arrow's speed just before hitting the cliff with height [tex]H[/tex], we have to find [tex]V[/tex] when [tex]y=H=21.612 m[/tex].
Using equation (2):
[tex]V^{2}={V_{o}}^{2} - 2gH[/tex] (9)
[tex]V=\sqrt{{V_{o}}^{2} - 2gH}[/tex] (10)
[tex]V=\sqrt{{(26 m/s)}^{2} - 2(9.8m/s^{2})(21.612 m)}[/tex] (11)
Finally:
[tex]V=15.887 m/s[/tex] (12) This is the arrow's impact speed
