Bird flies w/ speed of 12.2 over water, drops fish. a=gravity at 9.81. The alt. of bird=5.3. Air resistance=none. What is the bird speed at impact?
I'm sorry this was so condensed. The site permitted that I use only 150 characters.
I disagree with the above answer. It is correct about the vertical speed of the fish at impact. But then it goes on to add the effect of the horizontal speed of the bird. This is not relevant. I quote from a site on this topic:
This episode looks at the independence of vertical and horizontal motion. It concerns objects accelerating vertically when projected horizontally or vertically. The crucial concept is that vertical acceleration does not affect horizontal velocity. This explains all projectile motion.
So, the speed of the fish when it hits the water is 10.197 m/s^2. I got this by using two equations.
First, find the time the fish fell using
distance = .5 at2 where a is the acceleration and t is the time.
We know the distance and the acceleration so
5.3 = .5(9.8)t^2
t^2 = 1.0805 and t = 1.0394
So the fish fell for 1.0394 seconds.
Now we must use the formula for velocity:
v = at
We know acceleration (9.81) and we know time (1.0394) so that gives us
v = 10.197 m/s^2.
The above answer is only incorrect because it goes on to add in the effect of the horizontal motion, which it should not do.
BTW, you can put as much info as you want in the second box. So you could have said "please help with this physics problem" and then put the whole problem in the second box.
This is a question under the motion of a body in a projectile path.
The initial speed of the fish is =12.2 meter impatrted by the bird.
The acceleration due to gravity is 9.81 m/s^2.
The altitude at which the fish is dropped is 5.3 m
The initaial speed is 12.2 m/s in horizontal direction, and is unaffected by gravitational acceleration.
The vertical speed when the fish is dropped is zero and is continuously affected by the acceleration due to gravity. The final vertical speed of the dropped fish by the time it reaches water below 5.3m is given by:
v =sqrt(- u^2+2gH), where v is final vertical speed, u is the initial vertical speed of the dropped fish, g is the acceleration due to gravity and H is the height of the fish when it was dropped.
v = sqrt(-0^2+2*9.81*5.3)=10.1974m/s.
Since the bird has a horizontal unaffected velocity of 12.2m/s, the resulting velocity due to horizontal and vertical components of velocities is: sqrt(12.2^2+10.1974^2) m/s