An arrow is shot from a height of 1.55 m toward a cliff of height H. It is shot with a velocity of 26 m/s at an angle of 60° above the horizontal. It lands on the top edge of the cliff 3.99 s later. (a) Draw a sketch of the given example. Include the x-y coordinate system.
(b) What is the height of the cliff?
(c) What is the maximum height reached by the arrow along its trajectory?
(d) What is the arrow's impact speed just before hitting the cliff?

Explanation:

vertical component of the velocity of arrow

= 26 sin 60 = 22.516 m

height reached by it after 3.99 s

h = ut - 1/2 g t²

= 22.516 x 3.99 - .5 x 9.8 x 3.99²

= 89.83 - 78

11.83 m

Total height of cliff = 1.55 + 11.83

= 13.38 m

c ) maximum height covered s

v² = u² - 2gs

0 = u² - 2gs

s = u² / 2g

= 22.516² / 2 x 9.8

= 25.86

maximum height reached

= 25.86 + 1.55

= 27.41 m

d )

vertical speed after 3.99 s

v = u - gt

= 22.516 - 9.8 x 3.99

= -16.586

Horizontal component will remain unchanged

Horizontal component = 26 cos 60

= 13 m /s

Resultant of two velocities

= √ 13²+ 16.568²

= 21 m /s

the working equation for the angular size is the product of the physical size and 360 degrees divided by the product of two pi and distance from earth. the minimum size would be 0.559 degrees while the maximum size would be 0.49 degrees.