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 30.030.030.0 m below its starting point 5.005.005.00 s after it leaves the thrower's hand. Ignore air resistance. How high does it rise above its starting point?

Identify the known values: the displacement of the egg is 30.0 m downward, the time taken is 5.00 s, and the acceleration due to gravity is approximately 9.81 m/s² downward. Use the kinematic equation for displacement: s=$$v_{0}$t+12at^2$$, where s is the displacement, v_{0} is the initial velocity, t is the time, and a is the acceleration. Rearrange the equation to solve for the initial velocity v_{0}: v_{0}=$$s-12at^2t. $$Substitute the known values to find v_{0}. Determine the maximum height reached using the formula: v_{f}^2=v_{0}^2-2ah, where v_{f} is the final velocity (0 m/s at the peak), h is the height above the starting point. Solve for h using the rearranged equation: h=v_{0}^22a. Substitute the calculated initial velocity and known acceleration to find the height.

Ch 02: Motion Along a Straight Line

Young & Freedman Calc - University Physics 14th Edition

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Young & Freedman Calc 14th Edition

Ch 02: Motion Along a Straight Line

Problema 46b

Capitolo 2, Problema 46b

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. How high does it rise above its starting point?

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Identify the known values: the displacement of the egg is 30.0 m downward, the time taken is 5.00 s, and the acceleration due to gravity is approximately 9.81 m/s² downward.

Use the kinematic equation for displacement:

s = v

₀t+12at^2, where

s

is the displacement,

v

₀ is the initial velocity,

t

is the time, and

a

is the acceleration.

Rearrange the equation to solve for the initial velocity

v

₀:

v

₀=s-12at^2t. Substitute the known values to find

v

₀.

Determine the maximum height reached using the formula:

v

_f^2=v₀^2-2ah, where

v

_f is the final velocity (0 m/s at the peak),

h

is the height above the starting point.

Solve for

h

using the rearranged equation:

h = v

₀^22a. Substitute the calculated initial velocity and known acceleration to find the height.

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Kinematics Equations

Kinematics equations describe the motion of objects under constant acceleration. They relate displacement, initial velocity, final velocity, acceleration, and time. In this problem, these equations help determine the maximum height reached by the egg by analyzing its vertical motion, given the time and displacement.

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Percorso guidato

08:25

Kinematics Equations

Free Fall Motion

Free fall motion refers to the movement of an object under the influence of gravitational force alone. In this scenario, the egg is subject to Earth's gravity, which accelerates it downward at approximately 9.81 m/s². Understanding free fall is crucial to calculate how high the egg rises before gravity slows it to a stop.

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08:36

Vertical Motion & Free Fall

Initial Velocity

Initial velocity is the speed and direction of an object at the start of its motion. It is a critical factor in determining the egg's trajectory and maximum height. By knowing the initial velocity, one can use kinematic equations to find out how high the egg travels before gravity brings it back down.

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Percorso guidato

7:27

Escape Velocity

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