Posted by penny on Monday, September 13, 2010 at 11:13pm.
Use the equations
V = 5.1 - g t
Y = Yo + 5.1 t - (g/2) t^2
Yo is the elevation where the release occurs. You know what g is, I'm sure.
For (c), remember that the helicopter will be 10.2 m higher than Yo at t = 2 s, sincve it continues to rise. The bag will be lower than Yo.
part c is what I am having trouble with. what equation do I use?
delta Y (distance of bag below helicopter)
= Yhelicopter - Ybag
= Yo + 5.1 t - (g/2) t^2 - Yo -5.1t
= -(g/2)t^2
Related Questions
physics - A helicopter is rising at 5.5 m/s when a bag of its cargo is dropped...
physics - A helicopter is rising at 4.7 m/s when a bag is dropped from it. (...
physics - A helicopter is rising at 5.2 m/s when a bag of its cargo is dropped. ...
physics - A helicopter is rising at 5 m/s when a bag is dropped. After 2 seconds...
Physics - Kyle is flying a helicopter and it is rising at 4.0 m/s when he ...
Physics - Kyle is flying a helicopter and it is rising at 6.0 m/s when he ...
physics - A package is dropped from a helicopter that is descending steadily at ...
Physics - A bag is dropped from a hovering helicopter. The bag has fallen for 1....
Physics - A bag is dropped from a hovering helicopter. The bag has fallen for 2....
physics - A small mailbag is released from a helicopter that is descending ...
For Further Reading
