An egg is thrown nearly vertically upward from a point near the cornice of a tal

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

What is the initial speed of the egg? v=?m/s

How high does it rise above its starting point h=?m

What is the magnitude of its velocity at the highest point? v=?m/s

What is the magnitude of its acceleration at the highest point? a=?m/s^2

Explanation / Answer

Let's take vertically upward direction as positive and downward direction as negative

a) from the distance formula

X(t)=vt+(1/2)at2

Where x=-39 m and a=-g=9.8 m/s2

-39m=v(5)+(1/2)(-9.8m/sec2)(5sec)2

v(5)=112.5-39

v=83.5/5

v=16.7m/sec

So the initial speed of the egg is 16.7 m/sec

b)at highest point K.E=P.E (conservation of energy)

(1/2)mv2=mgh

h=(v2/2g)

=16.72/(2x9.8)

=14.23 m

Hight travelled by the egg is 14.23 m

c) at the highest point velocity v=0 m/sec

d) at the highest point a=g=9.8 m/sec2(downward)

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.