If a rock thrown upwards with a velocity 10m/s, it Height after t sec is given by H=10t-1.86t^2?

This is more of a math problem than a physics but its cool. I got you covered. Just take the derivative(v=10-3.72t) of the function of height(or position) and you will get the velocity of the object Then you plug in the 1 second into the equation and you get that the velocity equals 6.28m/s. Part 2, you plug the a into the derivative and get 10-3.72a. Part 3, you calculate when the height will equal zero and that gives you the time when it hits the surface. After some algebra, you get that the equation had two solutions. One at t=0 and one at t=10/1.86. The second value for t is the one you will use because t=0, the function starts so this is not going to be where the t hits the ground. The velocity it will hit the ground at is calculated easily by plugging t=10/1.86 into the equation. This gives the value that v=-10m/s. So the answers are a=6.28m/s, b=10-3.72a, c=10/1.86, d=-10m/s.