A seagull, ascending straight upward at 6.4 m/s, drops a shell when it is 14 m above the ground.?

Question

A seagull, ascending straight upward at 6.4 m/s, drops a shell when it is 14 m above the ground.

(a) What is the magnitude and direction of the shell's acceleration just after it is released?


(b) Find the maximum height above the ground reached by the shell.

(c) How long does it take for the shell to return to a height of 14 m above the ground?

(d) What is the speed of the shell at this time?

Answer 1

a) The acceleration is gravity -9.8 m/s

b) The kinetic energy will be converted to potential energy as
.5*m*v^2=m*g*h
where h is the height above the release point of 14 m

h=6.4^2/(2*9.81)

=2.09 m

max height is 16.09 m

c) The flight time up and back is twice the time to apogee
v(t)=0 at apogee
0=6.4-9.8*t
t=0.65
and the total flight time is 1.3 seconds

d) the speed is -6.4 m/s (downward velocity)

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