A few days ago
PHYSiCS HELP PLEASE!!!!!?
A pelican flying along a horizontal path drops a fish from a height of 3.6 m. The fish travels 6.8 m horizontally before it hits the water below.
The acceleration of gravity is 9.81 m/s^2.
a) What was the pelican’s initial speed? Answer in units of m/s.
b) If the pelican was traveling at the same speed but was only 3.0 m above the water, how far would the fish travel horizontally before hitting the water below? Answer in units of m.
I can’t figure out which equation to use to solve either part… solve/explain please??
Top 1 Answers
A few days ago
Favorite Answer
1) find the time between pelican dropped the fish and the fish hit the water. Use: initial vertical speed of th fish = 0, vertical acceleration = g = 9.81. So the height from which the pelican dropped the fish = h = t**2*g/2.
2) Now, assume that air resistance is zero, so the horizontal speed of the fish remains constant and equal to the horizontal speed of the pelican = v – unknown. You know the time that the fish travelled, you know the distance = d = 6.8. Use: d=v*t.
For (b) repeat all the above with the height equal 3.0
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