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 34.0 m below its starting point at a time 5.00 s after it leaves the thrower's hand. Air resistance may be ignored.
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?
E) What are the magnitude and direction of its acceleration at the highest point?

Answer 1

Answer:

A) 17.7 m/s

B) 15.98 m

C) Zero

E) 9.8 m/s²

Explanation:

given information

distance, h = - 34 m

time, t = 5 s

A) What is the initial speed of the egg?

h - h₀ = v₀t - $\frac{1}{2} g$t², h₀ = 0

- 34 = v₀ 5 - \frac{1}{2} 9.8 5²

- 34 = 5 v₀ - 122.5

v₀ = 122.5 - 34/5

    = 17.7 m/s

B) How high does it rise above its starting point?

v² = v₀² - 2gh

v = 0 (highest point)

2gh = v₀²

h = v₀²/2g

  = 17.7²/2 (9.8)

  = 15.98 m

C) What is the magnitude of its velocity at the highest point?

v = 0 (at highest point)

E) What are the magnitude and direction of its acceleration at the highest point?

g= 9.8 m/s², since the egg is moved vertically, the acceleration is the same as the gravitational acceleration.

Answer 2

Part A. The initial speed of the egg is 17.7 m/s.

Part B. The maximum height of the egg is 15.98 m.

Part C. The magnitude of the velocity at its highest point is 0 m/s.

Part E. The magnitude and direction of the acceleration in the upward direction is 9.8 m/s2.

Velocity and Acceleration

Given that the distance traveled by the egg is 34 m in time 5 seconds.

The distance at its initial point will be zero.

Part A

The initial speed of the egg can be calculated by the position equation of the egg.

$x = x_0 + v_0t + \dfrac {1}{2}at^2$

Where x is the initial position of the egg that is 0 m, x_0 is the final position that is 34 m in time t = 5 s, v_0 is the initial speed and a is the acceleration. Since the egg is thrown vertically, hence the acceleration of the egg is similar to the gravitational acceleration.

$0 = 34 + v\times 5 + \dfrac {1}{2}\times(- 9.8)\times 5^2$

$v = \dfrac {122.5 - 34}{5}$

$v = 17.7 \;\rm m/s$

The initial speed of the egg is 17.7 m/s.

Part B

The maximum height for the egg can be calculated by the equation given below.

$v^2 = v_0^2 + 2gh$

For the highest point, the final speed v will be zero and the initial speed v_0 is 17.7 m/s. Hence,

$0 = (17.7)^2 + 2\times (-9.8)\times h$

$19.6 h = 313.29$

$h = 15.98 \;\rm m$

The maximum height of the egg will be 15.98 m.

Part C

The velocity of the object thrown at a height will be zero at its highest point. Hence the velocity of the egg will be zero at its highest point.

Velocity $v_h$ = 0 m/s

Part E

The egg is thrown vertically upward, hence its acceleration is similar to the gravitational acceleration.

Acceleration a = 9.8 m/s2

To know more about the velocity and acceleration, follow the link given below.

https://brainly.com/question/2239252.