The path of a falling object?

Question

The path of a falling object is given by the function where represents the initial velocity in ft/sec and represents the initial height. The variable t is time in seconds, and s is the height of the object in feet.

a) If a rock is thrown upward with an initial velocity of 32 feet per second from the top of a 40-foot building, write the height equation using this information.
How is the rock after 0.5 seconds? After how many seconds will the rock reach maximum height? What is the maximum height?

Answer 1

height = v * t - 0.5 * g * t^2



v = initial velocity = 32 ft/s

g = acceleration of gravity

t = time in seconds

Answer 2

This question really is too easy to need asking.



y = y₀ + V₀ t + g t² / 2

g = -32 ft sec⁻²

V₀ = 32 ft sec⁻¹

y₀ = 40 ft



a) height at t=0.5 sec

t = 0.5 sec

y = 40 ft + (32 ft sec⁻¹)(0.5 sec) - (32 ft sec⁻²)(0.5 sec)² / 2

y = 52 ft



b) time of max height

dy/dt = V₀ + g t = 0

t = -V₀ / g

t = -32 ft sec⁻¹ / (-32 ft sec⁻²) = 1 sec



c) max height

y = 40 ft + (32 ft sec⁻¹)(1 sec) - (32 ft sec⁻²)(1 sec)² / 2

y = 56 ft



In case somebody asks when the rock will hit the ground at the bottom of the building...



y = A t² + B t + C = 0

A = -16

B = 32

C = 40

t = { -B ± sqrt (B² - 4AC) } / (2A)

The "minus" root is the useful one, since the "plus" root backtraces the trajectory before the rock was tossed upward.

t = 1 + sqrt(7/2)

t ≈ 2.87082869 sec

Answer 3

Gravity accelerates objects at 32 feet per second per second.



So if you throw it straight up at 32 feet per second, it will rise to 32 feet in one second, fall 32 feet in 1 second, and hit you.

Answer 4

this is perfectly straightforward calculation with equations you should already know. what's the problem?

Answer 5

Do your own homework.

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