Question

An egg is thrown nearly vertically upward from a point near the cornice of a tall building. The egg just misses the cornice on the way down and passes a point 30.0 m below its starting point 5.00 s after it leaves the thrower’s hand. Ignore air resistance.

(a) What is the initial speed of the egg?
(b) How high does it rise above its starting point?
(c) What is the magnitude of its velocity at the highest point?
(d) What are the magnitude and direction of its acceleration at the highest point?
(e) Sketch ay−t,vy−ta y −t,v y −t, and y-t graphs for the motion of the egg.

Answer 1

1. 18.5m/s

2. 17.5 m

3. 0 at its highest point

4. Direction is downwards

Step-by-step explanation:

1. This egg is thrown vertically from a height

Yo = 0. This egg then falls to the point y = -30.0 at t = 5seconds

Y-Yo = V0t - 1/2gt²

-30-0 = V0(5)-1/2(9.8)(5²)

-30 = 5v0 - 4.9x25

-30 = 5V0 - 122.5

-30+122.4 = 5v0

V0 = 92.5/5

= 18.5m/s

this is the initial speed of the egg

2. When the egg is at a maximum height it would have a velocity equal to 0

V² = V0² - 2*g*y

V = 0, V0 = 18.5, g = 9.8

0 = 18.5²-2x9.8*y

342.25-19.6y = 0

342.25 = 19.6y

Divide through by 19.6

Y = 342.25/19.6

Y = 17.5m

this value is how high it rises above starting point

3.

The magnitude of velocity is = 0 at its highest point

4.

This egg falls under gravity. Therefore the acceleration due to gravity has a constant magnitude and direction. Magnitude = 9.8m/s and it's direction is downwards.

5. Please check attachment for graph