- #1
cowgiljl
- 63
- 1
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.
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
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