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)

Question

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

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3 years 2021-08-27T08:06:42+00:00 1 Answers 7 views 0

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