Question

a stone is projected at a cliff of height h with initial speed of 42.0 m/s directed at an angle of 60 degrees above the horizontal. The stone strikes at A, 5.50s after launching. Find the height of the cliff, the speed of the stone just before the impact at A, and the max height H reached above the ground

Answer 1

Height of building = 51.81 meter

At point A

Speed in horizontal direction = 21 m/s

Speed in vertical direction = -17.585 m/s

Maximum height reached = 119.23 meter

Step-by-step explanation:

We have equation of motion ,
`s= ut+(1)/(2) at^2`, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

Vertical motion of stone ( up direction is positive):

Initial velocity = 42 sin 60 = 36.37 m/s, acceleration = -9.81
`m/s^2`, t = 5.50 seconds, we need to calculate s.

`s=36.37*5.5-(1)/(2) *9.8*5.5^2\\ \\ =51.81 meter`

So height of building = 51.81 meter

Speed of stone is divided in to 2 parts

Speed in horizontal direction remains constant = 42 cos 60 = 21 m/s

Speed in vertical direction

We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

u = 36.37 m/s, a = -9.81
`m/s^2` and t = 5.5 seconds

v = 36.37 - 9.81*5.5 = -17.585 m/s (towards down direction)

Speed in horizontal direction = 21 m/s

Speed in vertical direction = -17.585 m/s

We have equation of motion,
`v^2=u^2+2as`, where u is the initial velocity, u is the final velocity, s is the displacement and a is the acceleration.

At maximum height, v = 0 m/s, u = 36.37 m/s and a = -9.81
`m/s^2`

`0=36.37^2-2*9.81*s\\ \\ s=67.42 meter`

So maximum height reached = 67.42+51.81 = 119.23 meter.