A person standing on a cliff extends their arm past the cliff’s edge and throws

Question

A person standing on a cliff extends their arm past the cliff’s edge and throws a stone with a velocity 16.0 m/s downward. The stone is 80.0m above ground when it leaves the person’s hand.
A) Draw a diagram of the problem and label the known and unknown quantities
B) When will the stone land?
C) What is the stone’s velocity when it lands?
Show all work 2. A person standing on a cliff extends their arm past the cliffs edge and throws a stone with velocity 16.0 m/s downward. The stone is 80.0 m above ground when it leaves the person's hand a) Draw a diagram of the problem and label the known and unknown quantities. 80.0 b) When will the stone land? c) What is the stone's velocity when it lands? Show your work below:

Explanation / Answer

Assuming downward is positve and upward is negative.

Given Quantities are:

U = initial Velocity = +16 m/sec

V = final Velocity = ?

time taken to land, t = ?

a = acceleration = +g = 9.8 m/sec^2

Height of building = 80 m

Mark these quantities on figure

B.

Now Using equation

H = U*t + 0.5*a*t^2

80 = 16*t + 0.5*9.8*t^2

4.9*t^2 + 16t - 80 = 0

Solving above quadratic equation

t = [-16 +/- sqrt (16^2 + 4*4.9*80)]/(2*4.9)

take only positive root

t = 2.725 sec = 2.73 sec

Part C.

Using equation

V = U + a*t

V = 16 + 9.8*2.725

V = 42.705 m/sec = 42.7 m/sec

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