Solved Problems in Linear Motion – Up and down motion in free fall

1. A person throws a ball upward into the air with an initial velocity of 20 m/s. Calculate how high it goes. Ignore air resistance. Acceleration due to gravity (g) = 10 m/s^{2}.

Solution

We use one of these kinematic equations for motion at constant acceleration, as shown below.

v_{t} = v_{o} + a t

s = v_{o} t + ½ a t^{2}

v_{t}^{2} = v_{o}^{2} + 2 a s

__Known :__

*We choose the upward direction as positive and downward direction as negative.*

Initial velocity (v_{o}) = 20 m/s (positive upward)

Acceleration of gravity (g) = – 10 m/s^{2 }(negative downward).

Final velocity (v_{t}) = 0 (it’s speed is zero for an instant at highest point)

__Wanted :__ Maximum height (h)

__Solution :__

v_{t}^{2} = v_{o}^{2} + 2 g h

0 = (20^{2}) + 2(-10) h

0 = 400 – 20 h

400 = 20 h

h = 400 / 20 = 40 / 2 = 20 meters

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2. A person throws a stone upward at 20 m/s while standing on the edge of a cliff, so that the stone can fall to the base of the cliff 100 meters below.

(a) How long does it take the ball to reach the base of the cliff (b) Final velocity just before stone strikes the ground. Acceleration due to gravity (g) = 10 m/s^{2}. Ignore air resistance.

__Known :__

*We choose the upward direction as positive and downward direction as negative.*

High (h) = -100 meters (negative because final position below initial position)

Initial velocity (v_{o}) = 20 m/s (positive upward)

Acceleration of gravity (g) = -10 m/s^{2 }(negative downward)

__Wanted : __

(a) Time in air or time interval (t)

(b) Final velocity (v_{t})

__Solution :__

(a) Time interval (t)

__Known :__

High (h) = -100 meters (negative because final position below initial position)

Initial velocity (v_{o}) = 20 m/s (positive upward), Acceleration of gravity (g) = -10 m/s^{2} (negative downward).

h = v_{o} t + ½ g t^{2}

-100 = (20) t + ½ (-10) t^{2}

-100 = 20 t – 5 t^{2}

-5 t^{2} + 20 t + 100 = 0

We use quadratic formula :

(b) Final velocity

v_{t}^{2} = v_{o}^{2} + 2 g h

v_{t}^{2} = (20^{2}) + 2 (-10)(-100)

v_{t}^{2} = 400 + 2000

v_{t}^{2} = 2400

v_{t} = 49 m/s

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- Distance and displacement
- Average speed and average velocity
- Constant velocity
- Constant acceleration
- Free fall motion
- Down motion in free fall
- Up and down motion in free fall