Part A What is the initial speed of the egg? Part B How high does it rise above

Question

Part A

What is the initial speed of the egg?

Part B

How high does it rise above its starting point?

Part C

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

Part D

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

Part E

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

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 47.0 m below its starting point at a time 5.00 s after it leaves the throwers hand. Air resistance may be ignored

Explanation / Answer

A.

Vi = initial velocity = ?

a = -g = -9.81 m/sec^2

d = -47 m

t = 5 sec

Using equation

d = Vi*t + 0.5*g*t^2

-47 = Vi*5 - 0.5*9.81*5^2

Vi = (0.5*9.81*25 - 47)/5 = 15.13 m/sec

B.

Vi = 15.13 m/sec

Vmax = Velocity at highest point = 0 m/sec

a = -g

Using equation

Vmax^2 = Vi^2 + 2*a*Hmax

Hmax = (0^2 - 15.13^2)/(2*(-9.81))

Hmax = 11.67 m

C.

At the highest point Vmax = 0 for any vertical motion.

Vmax = 0 m/sec

D.

at the max point a = -g

|a| = 9.81 m/sec^2

E.

direction of acceleration is downward.

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