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 50.0 mm below its starting point at a time 5.00 ss after it leaves the thrower's hand. Air resistance may be ignored.

1. What is the initial speed of the egg?
v=?m/s
2. How high does it rise above its starting point
h=?m
3. What is the magnitude of its velocity at the highest point?
v=?m/s
4. What is the magnitude of its acceleration at the highest point?
a=?m/s^2

Answer 1

Step-by-step explanation:

Displacement ( downwards ) = 50 m , initial speed = - u (upwards ) ,

time = 5 s .

acceleration due to gravity( downwards) = 9.8 m /s² .

s = ut + 1/2 g t²

50 = - 5 u + .5 x 9.8 x 5²

50 = -5u + 122.5

5u = 122.5 - 50

u = 14.5 m /s

2 )

final velocity v = 0 , height upto which it rises = h

v² = u² - 2 g h

0 = 14.5² - 2 x 9.8 h

h = 10.72 m

3 )

At the highest point velocity = 0

4 )

At the highest point acceleration = 9.8 m /s² downwards .