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

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 .

all of the answers are correct.

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explanation:

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft