# Homework in college

cowgiljl
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

Homework Helper
a) is correct. The stone will have speed upward after t seconds of
14.7- 9.8t and will continue upwards until its speed is 0:
14.7- 9.8 t= 0 so t= 14.7/9.8= 1.5 sec. Since it is going downward after that, that is the highest point.

For (c)I get 11.025 meters but I won't quibble!

On (d) you've lost me completely. Did you drop (d) and miswrite (e) for (d)? (d) asks for speed when it hits the ground but you are calculating maximum height above ground.(Of course. Just add 19.6 to the answer to c.)

e) speed just as the stone hits the ground ("speed before it hits the ground" is ambiguous. Perhaps "JUST before it hits the ground"?):

You know, since you've already used it that
v(t)= 14.7- 9.8t and that h(t)= 19.6+ 14.7t- 4.9t2.

The stone will hit the ground when h(t)= 0 so solve
19.6+ 14.7t- 4.9t2=0 and put that value of t into
v= 14.7- 9.8 t. (There will be two times: one positive and one negative. Decide which to use!)
I get -63.7 m/s (the stone is going down at 63.7 m/s as it hits the ground.

Tyro
a) v = u + at
0 = 14.7 - 9.81(t)
t = 1.50s

S = ut + 1/2at^2
S = 14.7(1.5) - 0.5(9.81)(1.5^2)
= 22.05 - 11.03625
= 11.01375m

b) v^2 = u^2 + 2as
= 0 + 2(9.81)(11.01375+19.6)
v = 24.5ms-1

v = u + at
24.5 = 0 + 9.81(t)
t = 2.498s

Total time = 4.498s

c) Height = 11.01375 + 19.6 = 30.61375m

d) v = 24.5ms-1

Tyro
Damn, HallsofIvy you beat me to the answer BTW, are you sure of the 63.7ms-1 figure?

Unless I am reading the question wrong the stone is effectively 'dropped' with a zero initial velocity at its apex. The drop height is about 30m and without checking too much, common sense says 63.7 is a little high.

Homework Helper
You're right. By "conservation of energy", if it falls from 30 feet it should reach a speed of 25.5 m/s.