Finding An Initial Speed, Intro to Physics Question

[STEF2098]

A boy tosses a coin straight up to 2.8 m. Suppose the height when the coin leaves his hand is 1.2 m. What was the initial speed of the coin when it left his hand?

I'm just confused as to what equation to use. And I don't know whether to assume the acceleration is -9.8 m/s^2. Thanks.

 

[Matterwave]

You can do this either with forces, or with conservation of energy. The conservation of energy method is much simpler, but I'm not sure if you've learned it yet.

 

[kerimek]

I would first clarify that this scenario is assuming a constant gravitational acceleration of -9.8 m/s^2. This is a reasonable assumption for objects near the Earth's surface.

To find the initial speed of the coin, we can use the equation y = y0 + v0t + 1/2at^2, where y is the final height of the coin (2.8 m), y0 is the initial height (1.2 m), v0 is the initial speed (what we are trying to find), t is the time, and a is the acceleration (-9.8 m/s^2).

We can rearrange this equation to solve for v0: v0 = (y - y0)/t - 1/2at^2. Plugging in the given values, we get v0 = (2.8 m - 1.2 m)/t - 1/2(-9.8 m/s^2)t^2.

Since we do not have the value for time, we cannot solve for the initial speed without additional information. However, we can use this equation to find the initial speed at any given time during the coin's trajectory.

It is also important to note that this calculation assumes ideal conditions and does not take into account air resistance or other external factors that may affect the coin's motion.