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 42.0 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?

Explanation / Answer

Part A:

Use the following equation to find the initial velocity.

d = vi * t + ½ * a * t^2,

where, d represents the egg’s vertical displacement
Vertical displacement = final height – initial height = 0 – 42 = -42 meters
a = -9.8 m/s^2

put the values -

-42 = vi * 5 + ½ * -9.8 * 5^2
=> -42 = vi * 5 – 122.5
=> vi * 5 = 80.5
=> 80.5 / 5 = 16.1 m/s

Part B:
As it rises to its maximum height, its velocity decreases from 16.1 m/s to 0 m/s at the rate of 9.8 m/s^2.

So, use the following expression to determine the height -

vf^2 = vi^2 + 2 *a * d, vf = 0, a = -9.8 m/s^2

pu the values -
0 = 16.1^2 + 2 * -9.8 * d
=> 19.6 * d = 16.1^2
=> d = 16.1^2 ÷ 19.6 = 13.22 m

Part C:
Velocity at the heighest point shall be ZERO.

Part D:

Magnitude of acceleration at the highest point = 9.8 m/s^2

PartE:

Direction of acceleration at the highest point is in the DOWNWARD DIRECTION.

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