Acoin is tossed upward from an initial height of 7 m above the ground, with an initial speed of 7.4 m/s. the magnitude of the gravitational acceleration g = 9.8 m/s^2
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? note: speed has no sign, or always positive.

a) The coin´s maximum height is 9.79 m above the ground.

b) The coin is 2.17 s in the air.

c) The speed is 13.82 m/s when the coin hits the ground

Explanation:

The equations for the position and velocity of the coin are the following:

y = y0 + v0 · t + 1/2 · g · t²

v =  v0 + g · t

Where

y = height at time t

y0 = initial height

v0 = initial velocity

t = time

g = acceleration due to gravity

v = velocity at time t

a) At its max-height, the velocity of the coin is 0. Using the equation of velocity, we can obtain the time at which the velocity is 0.

v =  v0 + g · t

0 = 7.4 m/s - 9.8 m/s² · t

- 7.4 m/s / - 9.8 m/s² = t

t = 0.76 s

Now calculating the height of the coin at t = 0.76 s, we will obtain the maximum height:

y = y0 + v0 · t + 1/2 · g · t²

y = 0 m + 7.4 m/s · 0.76 s - 1/2 · 9.8 m/s² · (0.76 s)²

y = 2.79 m

The coin´s maximum height above the ground is 7 m + 2.79 m = 9.79 m

b) After the coin reaches its maximum height, it falls to the ground. The initial position will be the max-height (2.8 m) and the final position is the ground (-7 m). The initial velocity, v0, will be 0, because the coin is at the max-height. Then, using the equation of position we can calculate the time the coin is falling:

y = y0 + v0 · t + 1/2 · g · t²

-7 m = 2.79 m - 1/2 · 9.8 m/s² · t²

-2 ·(-7 m - 2.79 m)/ 9.8 m/s² = t²

t = 1.41 s

The coin is (1.41 s + 0.76 s) 2.17 s in the air

c) Using the equation of velocity, we can calculate the speed at time 1.41 s, when the coin hits the ground.

v =  v0 + g · t

v = 0 m/s - 9.8 m/s² · (1.41 s)

v = -13.82 m/s

The speed is 13.82 m/s when the coin hits the ground.

i don't know that i agree with that but what's the question ?

when a device is used in a circuit in which the voltage is 81 v, the current flowing through the device is 3 a. what is the resistance of the device?

a. 27 ω

b. 40.5 a

c. 243 ω

d. 243 a

the correct answer is a. 27