Question

An egg is thrown nearly vertically upward from a point near the cornice of a tall building. It just misses the cornice on the way down and passes a point a distance 41.0 m below its starting point at a time 5.00 s after it leaves the thrower's hand. Air resistance may be ignored.

1.What is the initial speed of the egg?

  m/s  

2.How high does it rise above its starting point?

  m

  

3.What is the magnitude of its velocity at the highest point?

  m/s  

4.What is the magnitude of its acceleration at the highest point?

  m/s2  

5.What is the direction of its acceleration at the highest point?

up

down

Answer 1

1)Let the maximum height reached by egg ,measured from cornice, be h

let initial speed be 'u'

time taken to reach maximum height =t₁= u/g

time taken to fall a height of h+41=5-t₁=5-(u/g)

h+41 = 0.5*g*[5-(u/g)]²...(i)

Also, u²=2*g*h=2gh

h=u²/2g...(ii)

Substituting this in (i), we get

(u²/2g)+41 = 0.5*g*[5-(u/g)]²=0.5*g*[25+u²/g² - 10u/g]=12.5g+0.5u²/g - 5u

Or, 41=12.5g-5u=122.5-5u

u=16.3 m/s

Initial speed pof egg = 16.3 m/s

2) Height point above starting point reached by egg = 16.3²/(2*9.8)=13.56m

3)at highest point, velocity is zero

4)Magnitude of accleeration throughout the entire trajectory(including highest point) is =acceleration due to gravity=g=9.8m/s²

5)Direction of accleeration throughout the entire trajectory(including highest point) is vertically downwards