This equations of motion under constant acceleration example problem shows how to determine the maximum height, velocity and time of flight for a coin flipped into a well. This problem could be modified to solve any object tossed vertically or dropped off a tall building or any height. This type of problem is a common equations of motion homework problem.

**Problem:**

A girl flips a coin into a 50 m deep wishing well. If she flips the coin upwards with an initial velocity of 5 m/s:

a) How high does the coin rise?

b) How long does it take to get to this point?

c) How long does it take for the coin to reach the bottom of the well?

d) What is the velocity when the coin hits the bottom of the well?

**Solution:**

I have chosen the coordinate system to begin at the launch point. The maximum height will be at point +y and the bottom of the well is at -50 m. The initial velocity at launch is +5 m/s and the acceleration due to gravity is equal to -9.8 m/s^{2}.

The equations we need for this problem are:

1) y = y_{0} + v_{0}t + ½at^{2}

2) v = v_{0} + at

3) v^{2} = v_{0}^{2} + 2a(y – y_{0})

**Part a)** How high does the coin rise?

At the top of the coin’s flight, the velocity will equal zero. With this information, we have enough to use equation 3 from above to find the position at the top.

v^{2} = v_{0}^{2} – 2a(y – y_{0})

0 = (5 m/s)^{2} + 2(-9.8 m/s^{2})(y – 0)

0 = 25 m^{2}/s^{2} – (19.6 m/s^{2})y

(19.6 m/s^{2})y = 25 m^{2}/s^{2}

y = 1.28 m

**Part b)** How long does it take to reach the top?

Equation 2 is the useful equation for this part.

v = v_{0} + at

0 = 5 m/s + (-9.8 m/s^{2})t

(9.8 m/s^{2})t = 5 m/s

t = 0.51 s

**Part c)** How long does it take to reach the bottom of the well?

Equation 1 is the one to use for this part. Set y = -50 m.

y = y_{0} + v_{0}t + ½at^{2}

-50 m = 0 + (5 m/s)t + ½(-9.8 m/s^{2})t^{2}

0 = (-4.9 m/s^{2})t^{2} + (5 m/s)t + 50 m

This equation has two solutions. Use the quadratic equation to find them.

where

a = -4.9

b = 5

c = 50

t = 3.7 s or t = -2.7 s

The negative time implies a solution before the coin was tossed. The time that fits the situation is the positive value. The time to the bottom of the well was 3.7 seconds after being tossed.

**Part d)** What was the velocity of the coin at the bottom of the well?

Equation 2 will help here since we know the time it took to get there.

v = v_{0} + at

v = 5 m/s + (-9.8 m/s^{2})(3.7 s)

v = 5 m/s – 36.3 m/s

v = -31.3 m/s

The velocity of the coin at the bottom of the well was 31.3 m/s. The negative sign means the direction was downward.

If you need more worked examples like this one, check out these other constant acceleration example problems.

Equations of Motion – Constant Acceleration Example Problem

Equations of Motion – Interception Example Problem

Projectile Motion Example Problem

Seems logical that the t=-2.7 seconds solution in part c describes when the coin would be at the bottom of the well had it been launched upward from the bottom of the well.