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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)
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Answer:
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