Question

football game customarily begins with a coin toss to determine who kicks off. The referee tosses the coin up with an initial speed of 5 m/s. In the absence of air resistance, how high does the coin go above its point of release? Hint: Draw the co-ordinate system. Note that the object is moving upwards, but the acceleration is downwards. They will have opposite signs, unline the rock in the well problem done in class where the rock fell in the downward direction and the velocity and acceleration were in the same direction.

Answer 1

Answer:

1.28 m

Explanation:

You are given the initial speed that is 5 m/s. Since the coin is sent upwards, this means that the speed is positive according to the co-ordinate system in the y axis. Additionally, there is the gravity that curbs the coin in the opposite direction making the final speed becomes zero at the highest point, so the gravity takes negative sign.

So, from the explanation we have:

$Vo = 5 m/s$

$V = 0 m/s$

$a = -9.8 m/s^2$

so, we use the following equation to find y (hight)

$V^2 = Vo^2 + 2ay$

$y = V^2 - Vo^2 / 2a$

$y = (0 m/s)^2 - (5 m/s)^2 / 2*(-9.8 m/s^2)$

$y = - (25 m^2 / s^2) / -19.6 m /s^2$

$y = 1.28 m$