A stone is thrown vertically upward with a speed of 28.0 m/s how much time is required to reach this height

A stone is thrown vertically upward with a speed of 17.0 m/s. How fast is it moving when it reaches a height of 11.0 m? How long is required to reach this height?

Let’s review the 4 basic kinematic equations of motion for constant acceleration (this is a lesson – suggest you commit these to memory):

s = ut + ½ at^2 …. (1)

v^2 = u^2 + 2as …. (2)

v = u + at …. (3)

s = (u + v)t/2 …. (4)

where s is distance, u is initial velocity, v is final velocity, a is acceleration and t is time.

In this case, we know u = 17.0m/s, a = -g = -9.81m/s^2, s = 11.0m and we want to know v and t, so from equation (2):

v^2 = u^2 + 2as

v^2 = 17.0^2 -2(9.81)(11.0)

v = √73.18 = 8.55m/s

now from equation (3):

v = u + at

8.55 = 17.0 – 9.81t

t = (8.55 – 17.0)/(-9.81) = 0.86s

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