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?

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 - 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.

(a) reaction at each front wheel is 5272N (upward)

(b) force between boulder and pallet is 4124N (compression)

Explanation:

Acceleration of the truck = 1 m/  (to the left)

when the truck moves 1 m to the left, the boulder is B and pallet A are raised 0.5 m, then,

=  0.5 m/ (upward) , =  0.5 m/ (upward)

Let T be tension in the cable

pallet and boulder: ∑fy = ∑(fy)eff = 2T- ( + )g =  ( + )

                                  = 2T- (400 + 50)*(9.81 m/) = (400 + 50)*(0.5 m/)

                        T = 2320N

Truck:   = ∑()eff: = (3.4m) + (2.0m) - T (0.6m)= (1.0m)

Nf = (2.0m)(2000 kg)(9.81 m/ )/3.4m -  (0.6 m)(2320 N)/3.4m + (1.0 m)(2000 kg)(1.0 m/)  = 11541.2N - 409.4N - 588.2N = 10544N

∑fy (upward) = ∑(fy)eff: + - g = 0

                                       10544 +  - (2000kg)(9.81 m/ ) = 0

                = 9076N

   ∑fx (to the left) = ∑(fx)eff:   - T =

                                      = 2320N + (2000kg)(9.81 m/ ) = 4320N

(a) reaction at each front wheel:

1/2 (upward): 1/2 (10544N) = 5272N (upward)

(b) force between boulder and pallet:

∑fy (upward) = ∑(fy)eff: + g -

            = (400kg)(9.81 m/) + (400kg)(0.5 m/) = 4124N (compression)