An object is thrown upward with an initial velocity of 50 feet per second from a building 35 feet tall. A)State and interpret the vertex. B)When will the object hit the ground? slader

a) The vertex of the object's motion measured from the top of the 35 ft tall building from which it was thrown = 38.82 ft.

But the vertex of the object's motion, measured from the ground = 73.82 ft

b) 3.694 s after it was thrown.

Explanation:

The vertex of an object's motion is the highest vertical distance travelled.

Using the equations of motion,

u = initial velocity of the object = 50 ft/s

v = velocity of the object at maximum height reached (the vertex) = 0 ft/s

y = highest vertical height reached = ?

y₀ = initial height of the ball = 35 ft

g = acceleration due to gravity = -32.2 ft/s²

v² = u² + 2g(y - y₀)

0² = 50² + (2)(-32.2)(y - 35)

(y - 35) = 2500/(2×32.2)

y - 35 = 38.82 ft

y = 38.82 + 35 = 73.82 ft

b) For the time taken for the object to hit the ground.

I will calculate the time taken to reach maximum height and add it to the time to reach the ground from maximum height.

Time taken to reach maximum height

u = initial velocity of the object = 50 ft/s

v = velocity of the object at maximum height reached (the vertex) = 0 ft/s

g = acceleration due to gravity = -32.2 ft/s²

t₁ = ?

v = u + gt

0 = 50 + (-32.2)t

t₁ = (50/32.2)

t₁ = 1.553 s

Time taken to reach the ground from maximum height

u = initial velocity of the object at maximum height = 0 m/s

t₂ = ?

y = 73.82 ft

g = acceleration due to gravity = 32.2 ft/s²

y = ut + (1/2)gt₂²

73.82 = 0 + (1/2)(32.2)t₂²

t₂² = (2×73.82)/32.2

t₂² = 4.585

t₂ = 2.141 s

T = t₁ + t₂ = 1.553 + 2.141

T = 3.694 s

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