Question

A coin is tossed upward from an initial height of 6 m above the ground, with an initial speed of 9.6 m/s m/s.

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

Take the point of release to be y0 = 0. Choose UPWARD as positive y direction.

Pay attention to the signs of position, velocity and acceleration.

Keep 2 decimal places in all answers.

(a) Find the coin’s maximum height in meters above the ground?

(b) How long in seconds is the coin in the air?

You need to set up a quadratic equation with time t. Solve it for time t. Only take the positive solution.

(c) What is its speed in m/s when it hits the ground?

Answer 1

a)

y_o = initial position of coin = 0 m

y_max = position of the coin at highest point = ?

v_o = initial velocity of launch = 9.6 m/s

v = final velocity at the highest point = 0 m/s

a = acceleration due to gravity = - 9.8 m/s²

Using the kinematics equation

v² = v_o² + 2 a (y_max - y_o)

0² = 9.6² + 2 (- 9.8) (y_max - 0)

y_max = 4.7 m

h = height of the point of launch above the ground = 6 m

H = height of point of maximum height above the ground

Maximum height above the ground is given as

H = y_max + h

H = 4.7 + 6

H = 10.7 m

b)

y_o = initial position of coin = 0 m

y = final position of the coin at ground = - 6 m

v_o = initial velocity of launch = 9.6 m/s

t = time of travel in air

a = acceleration due to gravity = - 9.8 m/s²

Using the kinematics equation

y = y_o + v_o t + (0.5) a t²

- 6 = 0 + (9.6) t + (0.5) (- 9.8) t²

t = 2.5 sec

c)

v_o = initial velocity of launch = 9.6 m/s

t = time of travel in air = 2.5 sec

a = acceleration due to gravity = - 9.8 m/s²

v = final velocity as it hits the ground

Using the equation

v = v_o + a t

v = 9.6 + (- 9.8) (2.5)

v = - 14.9 m/s

So speed at the time hitting the ground is 14.9 m/s