Coin flipped from height of 1.68 m. After 1.15 s, it lands 10.5 m away. With what speed did it leave the hand.

Teacher gave me the answer as 9.88 m/s. classmate and I keep getting 10.037 m/s. help. I don't think the answers are close enough to be considered a rounding error. Help?

-1.68 = Viy*t - ½g*t² → (using g = 9.8 m/s²)
Vyi = 4.174 m/s

Vix = D/t = 9.13 m/s


V = √[Vyi² + Vix²] = 10.039 m/s

Stick up for your correct answer

If we neglect air resistance (which is probably a decent assumption for this problem), this becomes a basic ballistics problem. The nice thing about problems like this is, there's only one force - gravity, and it only acts in one direction - vertically. So the horizontal motion can be treated independently from the vertical motion.

In the vertical direction, we know the coin experiences an acceleration of -10 m/s^2 (notice that we just defined positive vertical to be upward). It starts at a height of 1.68 m, and it takes 1.15 s to hit the ground. From that, we should be able to figure out the initial vertical speed. The equation of motion is

d(t) = d0 + v0 * t + a * t^2 / 2

where d(t) is the position as a function of time, d0 is the initial position, v0 is the initial velocity, and a is the constant acceleration. We know d0, we know a, and we know that at T = 1.15 s, d(T) = 0. So, what's v0?

v0 = (d(T) - d0 - a * T^2 / 2) / T

Plugging in values, we get

v0 = (0 - 1.68 - (-10) * 1.15^2 / 2) / 1.15
v0 = 4.289 m/s

That's the initial vertical velocity. Note that it's positive, so it's upward. Also note, I'm not using units in my calculation, but you should.

Now, let's look at the horizontal motion. This is simple. Since there are no horizontal forces, there's no horizontal acceleration, so the t^2 term drops out of the constant acceleration equation of motion.

d(t) = d0 + v0 * t + 0 * t^2 / 2
d(t) = d0 + v0 * t + 0
d(t) = d0 + v0 * t