Answer:t=0.3253 s
Explanation:
Given
speed of balloon is 
speed of camera 
Initial separation between camera and balloon is 
Suppose after t sec of throw camera reach balloon then,
distance travel by balloon is


and distance travel by camera to reach balloon is


Now






There are two times when camera reaches the same level as balloon and the smaller time is associated with with the first one .
(b)When passenger catches the camera time is 
velocity is given by



and position of camera is same as of balloon so
Position is 

The sun is bright and when its cold there is no sun
Using the Equation:
v² = vi² + 2 · a · s → Eq.1
where,
v = final velocity
vi = initial velocity
a = acceleration
s = distance
<span><span>We know that vi = 0 because the ball was at rest initially.
</span><span>
Therefore,
Solving Eq.1 for acceleration,
</span></span> v² = vi² + 2 · a · s
v² = 0 + 2 · a · s
v² = 2 · a · s
Rearranging for a,
a = v ²/2·<span>s
Substituting the values,
a = 46</span>²/2×1<span>
a = 1058 m/s</span>²
<span>Now applying Newton's 2nd law of motion,
</span>
<span>F = ma
= 0.145</span>×<span>1058
F = 153.4 N</span>