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

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?

Answer 1

A) 15.3 m/s

B) 11.94m

C) 0 m/s

D) 9.8 m/s²

E) Direction is downwards

Step-by-step explanation:

A) If we take the origin at the cornice of the building and the positive y direction to be upward, we will arrive at the equation;

y - y_o = ut - (1/2)gt²

From the question, y_o = 0 m, y = 46 m and t = 5 s

Thus, we have;

-46 = 5u - (1/2)(9.8)(5²)

-46 = 5u - 122.5

5u = 122.5 - 46

u = 76.5/5 = 15.3 m/s

B) when the egg is at a maximum height, its final velocity there is zero. Thus using the equation ;

v² = u² - 2gy

So, 0² = 15.3² - 2(9.8y)

0 = 234.09 - 19.6y

So, 19.6y = 234.09

y = 234.09/19.6 = 11.94m

C) Magnitude of velocity at highest point is zero

D) At the highest point, the egg falls under the effect of gravity. Thus the acceleration due to gravity is constant and the value remains 9.8 m/s²

E) At the highest point, the egg falls under the effect of gravity. Thus the direction of acceleration due to gravity is constant throughout and the direction from this point is downwards.

Answer 2

Step-by-step explanation:

Given that,

Height = 46m

Then, final height is y=-46m

Initial height yo=0m

Time to reach height is 5sec

a. Initial speed

Using equation of motion

y-yo = vot - ½gt²

-46-0 = vo• 5- ½×9.8×5²

-46 =5vo -122.5

5vo = -46+122.5

5vo =76.5

vo = 76.5/5

vo= 15.3m/s upward.

vo = 15.3m/s

b. How high does it rise

Using equation of free fall

At maximum height, the final velocity is 0m/s

v² = u² - 2gH

0² = 15.3² -2•9.8H

-15.3² = -19.6H

H=15.3²÷19.6

H=11.943m

Maximum height 11.943m

c. At maximum height the velocity is zero this is due to gravity provides the same acceleration to the ball on the way up (slowing it down) as on the way down (speeding it up), the time to reach maximum altitude is the same as the time to return to its launch position.

d. The magnitude of acceleration at the highest point is the same as the acceleration due to gravity(g=9.8m/s²).

e. The direction of the acceleration is downward, it is directed toward the negative y axis( -j)

a = -9.81m/s².