Astone is thrown up vertically upward from the edge of a building 19.6 meters high with an inital velocity of 14.7 m/sec. The stone just misses the building on the way down and strikes the street below.(adsbygoogle = window.adsbygoogle || []).push({});

A) time to reach max height

b) total time of flight

C) height above the building

d) the velocity of the stone before it hits the ground

I think I am on the right track could anybody tell me if they are right?

A) 0 - 14.4m/se / -9.80 Where the time to the max height 1.5 sec

c) Ymax = Vo(t) + .5(-g)(t squared) the formula use

Ymax = 22.05 -11.03

Ymax is 11.0 meters above the building

D) Height above ground = building height + stone max height

19.6m + 11.0m = 30.6 m

B) 2h/g = t squared Which Tdown is = 2.5 sec and Tup is 1.5 secand get 4.0 secs.

Should this be more time becaues the stone would have been in flight for 3 sec. When it becomes = with the build on the way down before accel more?

E) Vf = g (Tdown) Vf = -9.80 (2.5)

Vf is = to -24.5 m/sec

Thanks you

Joe

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