Question

A lead ball is thrown vertically downward with an initial speed of 3.9 m/s from the top of a building, which is 105 m high.

(Compare this situation to: A ball is dropped from rest from the top of a building, which is 105 m high. )

The magnitude of the gravitational acceleration g = 9.8 m/s2

Keep 2 decimal places in all answers.

In this problem, the following setup is convenient:

Take the initial location of ball (the top of the building) as origin x0 = 0

Take DOWNWARD as +x)

(d) How long (in seconds) does it take the ball to hit the ground?

You need to solve a quadratic equation for the time t. There are two solutions.

What is the negative solution (of time) to your quadratic equation?

s

(5 attempts remaining)

What is the positive solution (of time) to your quadratic equation?

s

Answer 1

Part D.

Using 2nd kinematic equation:

d = U*t + (1/2)*a*t^2

As given in question, we've assumed that downward direction is +ve

d = vertical distance travelled by ball when it reaches at ground = 105 m

U = Initial speed of ball = 3.9 m/sec

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

Using above values:

105 = 3.9*t + (1/2)*9.8*t^2

4.9*t^2 + 3.9*t - 105 = 0

Now solving above quadratic equation:

t = [-3.9 sqrt (3.9^2 - 4*(-105)*4.9)]/(2*4.9)

t = [-3.9 sqrt (2073.21)]/(9.8)

taking -ve sign

t1 = negative solution of quadratic equation

t1 = [-3.9 - sqrt (2073.21)]/(9.8)

t1 = -5.04 sec

taking +ve sign

t2 = positive solution of quadratic equation

t2 = [-3.9 + sqrt (2073.21)]/(9.8)

t2 = 4.25 sec

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