Question

A camera falls from a blimp that is 492 m
above the ground and rising at a speed of
8.49 m/s.
Find the maximum height reached by the
camera with respect to the ground. The acceleration
of gravity is 9.8 m/s
2
.
Answer in units of m.
007 (part 2 of 2) 10.0 points
Find the speed at which it hits the ground.
Answer in units of cm/s.

PLS HELP AND GIVE ANSWERRR

Answer 1

a(0) = -9.8 m/s^2 
v(0) = 8.49 m/s 
s(0) = 492 m 

a(t) = -9.8 
v(t) = -9.8 * t + C 
v(0) = 8.49 
8.49 = -9.8 * 0 + C 
8.49 = C 
v(t) = -9.8 * t + 8.49 
s(t) = -4.9 * t^2 + 8.49 * t + C 
s(0) = 492 
492 = C 
s(t) = -4.9 * t^2 + 8.49 * t + 492 

v(t) = 0 
0 = -9.8 * t + 8.49 
9.8 * t = 8.49 
t = 8.49 / 9.8 
t = 849 / 980 

s(849/980) => 
-4.9 * (849/980)^2 + 8.49 * (849/980) + 492 => 
495.67755612244897959183673469387755102... 

It rises to 495.68 meters, roughly, then falls

Answer 2

The velocity of blimp is 8.49m/s.

The camera falls with zero velocity with respect to blimp.

So the velocity of camera wrt ground =-8.49m/s

Now the acceleration of gravity is 9.8m/s², so it will be taken negative.

By formula

v²=u²+2as

u=+8.49m/s

a=-9.8m/s²

v=0 [speed at maximum height =0]

0²=(8.9)² -2(9.8)s

8.9²=19.6s

79.21=19.6s

s=79.21/19.6

=4m

The maximum height =492 +4=496m

Now , when camera is at highest position, initial velocity=0

acceleration =9.8m/s²

s=496m

v²=0-2(-9.8)(496)

v²=19.6×496

v²=9721.6

v=98.6 in downword direction