Question

A coin is tossed upward from an initial height of 9 m above the ground, with an initial speed of 8.5 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?

Answer 1

Answer:

h= 12.68 m  : Maximum height that the coin reaches from the ground.

Explanation:

Kinematic equations to describe the movement of the coin :

In vertical upward movement, the acceleration due to gravity makes sense opposite to displacement, then, g is negative in the equations:

Look at the attached graphic

Vf = Vo - g*t  Formula (1)

Vf² = Vo² - 2*g*y  Formula (2)

Vf : Final speed , Vf=0 at position y

Vo : Initial speed , Vo=8.5 m/s (positive)

g : Acceleration due to gravity , g = 9,8 m/s² (negative)

y: Position to Vf=0 , from an initial height of 9 m above the ground

Calculation of the time (t) in which the currency reaches the height y :(Look at the attached graphic)

We replace data in formula (1)

0=8.5 - 9.8*t

9.8*t = 8.5

t = 8.5/9.8=0.867 s

Calculation of y for the time t at which the final velocity (Vf) is equal to zero:

We replace data in formula (2)

0 = 8.5² - 2*9.8*y

2*9.8*y= 8.5²

y= 8.5²/19.6 =3.68m

Calculation of the maximum height (h) that the coin reaches

(Look at the attached graphic)

h= 9+y

h= 9+3.68 = 12.68 m