A seagull, ascending straight upward at 5.20 m/s, drops a shell when it is 15.0 m above the ground.
(a) What is the magnitude of the shell's acceleration just after it is released?
______ m/s^2
(b) Find the maximum height above the ground reached by the shell.
______ m
(c) How long does it take for the shell to reach the ground?
______ s
(d) What is the speed of the shell at this time?
_______ m/s
2007-09-24 15:58:54 · 3 answers · asked by acousticplaya90 1 in Science & Mathematics ➔ Physics
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2007-09-28 10:01:31 · answer #1 · answered by ? 6 · 0⤊ 0⤋
Momentum=mass x velocity So in basic terms numerous the products with the aid of the cost wherein it is going. working occasion, to discover the momentum of question type a million, numerous the mass of the rifle with the aid of the bullet's velocity. the something of the countless themes makes use of the comparable recommendations-set, so in basic terms repeat that recommendations-set.
2016-10-19 21:18:28 · answer #2 · answered by ? 4 · 0⤊ 0⤋
I will ignore air resistance
a) -9.81 m/s^2
b) v(t)=5.2-9.81*t
apogee is reached when v(t)=0
t=5.2/9.81
plug this into
y(t)=15+5.2*t-.5*9.81*t^2
y(t)=15+5.2^2/9.81
-.5*5.2^2/9.81
=15+.5*5.2^2/9.81
=16.38 meters
c)the shell hits the ground when y(t)=0
or
0=15+5.2*t-.5*9.81*t^2
solve for the positive root of t
2.357 seconds
d) plug 2.357 into v(t)
=5.2-9.81*2.357
-17.9 m/s
j
2007-09-25 05:57:07 · answer #3 · answered by odu83 7 · 1⤊ 0⤋