A stone is thrown vertically upward with a speed of 18.0. (a) How fast is it moving when it reaches a height of 11.0? (b) How long is required to reach this height?

For part A) I did: y=11m, g=9.81m/s^2, Vy0 = 18 m/s, y0=0, (V^2) y=?

V^2 y = V^2 y0 - 2g (y-y0)

--> V^2 y=18^2m/s-2 (9.8) (11m-0m)

--> Vy=Square root (324m/s-215.6) = 10.4m/s

Part B) I started out with:

y=y0 + Vy0t - (1/2) gt^2

--> 11 = 18t m/s - (1/2) 9.8t^2

--> - 11 + 18t - 9.8t^2

Question

Righty

Physics

20 September, 03:10

A stone is thrown vertically upward with a speed of 18.0. (a) How fast is it moving when it reaches a height of 11.0? (b) How long is required to reach this height?

For part A) I did: y=11m, g=9.81m/s^2, Vy0 = 18 m/s, y0=0, (V^2) y=?

V^2 y = V^2 y0 - 2g (y-y0)

--> V^2 y=18^2m/s-2 (9.8) (11m-0m)

--> Vy=Square root (324m/s-215.6) = 10.4m/s

Part B) I started out with:

y=y0 + Vy0t - (1/2) gt^2

--> 11 = 18t m/s - (1/2) 9.8t^2

--> - 11 + 18t - 9.8t^2

+1

Answers (1)

Know the Answer?

Mason · Answer 1

For the first part, we are looking for Vf when dy=11.0

Upward is positive, downward is negative.

So Vf = square root [2 (-9.8) (11.0) + (18.0) ^2]

Vf = 10.4 m/s your answer is correct.

For the part b, t is equals to the time took to reach and dy is equals to 11.0

you did, 11 = 18t m/s - (1/2) 9.8t^2 then - 11 + 18t - 9.8t^2. By quadratic formula, for the way down the answer is 2.9 s while on it's way up, the answer is 0.77 s