Solve Motion Problem: 50m Dropped Camera in 2.0m/s Descent

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The discussion revolves around a physics problem involving a hot air balloon descending at 2.0 m/s and a camera dropped from 50 m above the ground. The user initially attempted to solve for the time it takes for the camera to reach the ground using the equation 50 = (-2.0 m/s)t + (1/2)(-10 m/s²)t² but arrived at an incorrect time of 2.9 seconds. Participants suggested using a simpler kinematic formula that relates initial velocity, distance, and acceleration due to gravity to find the correct time and final velocity just before landing.

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can someone help me with this problem? A hot air balloon is descending at a rate of 2.0 m/s when a passenger drops a camera. (a) If the camera is 50 m above the ground when it is dropped, how long does it take to reach the ground? (b) What is its velocity just before it lands? Let upward be the positive direction for this problem.

i did part a by this formula. 50 = (-2.0m/s)t + (1/2)(-10m/s2)t2
s2 = s squared t2 = t squared
i tried to solve for t and i got 2.9, but that not the right answer. anyone know what i did wrong?
 
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oh sorry i just saw the sticky for the homework forum. i'll post it there
 
Rather than forcing yourself to solve a quadratic equation, try using another simpler kinematic formula. You know initial velocity, distance, and acceleration due to gravity, and you need to know time: what formula relates all those variables?
 
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