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

Answer 1

Answer:

The initial speed of the egg is Vo = 16.7m/s

The highest point the egg gets to above the starting point is 14.2m

The magnitude of the veloc

Explanation:

The initial speed of the egg was computed by first setting the initial positionbof the egg to be y0 = 0 and then using the equivalent

y = y0 + Vo *t -1/2 * gt²

and solving for Vo

The highest point was calculated by first composure the time it took the egg to get the maximum height using the fact that when the egg gets to the highest point above the starting point it will temporarily be stationary having a zero velocity (v = 0). Taking advantage of this and setting v = to zero in the equation

V = Vo - gt

Where g is the acceleration due to gravity

Then using this equation again

Detailed solution can be found below in the attachment.

The acceleration of the egg is still the same as the acceleration due to gravity because no other force is acting on it and as a result it accelerates in the direction of gravity downloads. This answer is backed by Newton's second law of motion .

Thank you very much and I hope this is helpful to you.

Answer 2

The initial speed of the egg thrown from the given height is 16.7 m/s.

The maximum height traveled by the egg is 14.23 m.

The magnitude of its velocity at the highest point is zero.

The magnitude of its acceleration at the highest point is zero.

The given parameters;

• distance traveled, d = 39 m
• time of motion, t = 5.0 s

The initial speed of the egg is calculated as;

$-h = v_0_yt - \frac{1}{2} gt^2\\\\-39 = 5v_0_y - (0.5\times 9.8)(5)^2\\\\-39 = 5v_0_y - 122.5\\\\5v_0_y = 83.5 \\\\v_0_y = \frac{83.5}{5} \\\\v_0_y = 16.7 \ m/s$

The maximum height traveled by the egg is calculated as;

$v^2 = u^2 -2gh\\\\2gh = u^2 - v^2\\\\h = \frac{u^2 - v^2}{2g}$

where;

• v is the final velocity of the egg at the maximum height = 0

$h = \frac{(16.7)^2 }{2\times 9.8} \\\\h = 14.23 \ m$

The vertical  acceleration at the highest point is zero because the egg is under the influence of gravity directed downwards.

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