Answer:
The height of the cliff is 39.655 m
Given:
Height at which the arrow was shot, h = 1.3 m
Velocity of the arrow, u = 34 m/s
Angle, [tex]\theta' = 58.1^{\circ}[/tex]
Time of the fight, t = 4.1 s
Solution:
Let the Height of the cliff be H
Since, the motion of the object is projectile motion and the direction of motion is vertical at some angle
Therefore, we consider the vertical component of velocity, [tex]u_{y} = usin\theta[/tex].
Now,
The Height of the cliff is given by applying the second equation of motion in the projectile:
Thus
[tex]s = u_{y}t - \frac{1}{2}gt^{2}[/tex]
[tex]s = 34sin60^{\circ}\times 4.1 - \frac{1}{2}\times 9.8\times 4.1^{2}[/tex]
s = 38.355 m
Now, the height of the cliff, H:
H = s + h = 38.355 + 1.3 = 39.655 m
