1) A 155g baseball is dropped from a tree 15.0m above the ground.
A) If it actually hits the ground with a speed of 7.00m/s , what is the magnitude of the average force of air resistance exerted on it?
B) What is the direction of the average force of air resistance exerted on baseball?
A) Upward B) Downward C) Horizontally ( PLEASE CHOOSE ONE ANSWER)
PLEASE EXPLAIN!
JullyWum
Favorite Answer
By conservation of energy ..
PE lost (mgh) = KE gained (½mv²) + work against friction (F x h)
F x h = m(gh - ½v²)
F = m(gh - ½v²)/h .. 0.155kg (9.80*15m - ½*7.0²) / 15 .. ►F = 1.27 N
(B) Upwards .. against the downward motion.
Caroline
A) Draw a free-body diagram, and you'll see that some of the weight force will cancel with air resistance (drag) force. If you take the initial velocity to be zero, the final to be 7, and the y distance to be 15, use vf^2 = vi^2 + 2*a*y to solve for acceleration. Since F=ma, you can find the net force on the object.
This F is the net force on the ball, which is the weight force minus the drag force. Weight is always m*g on Earth, so you can solve for drag.
B) Drag force always opposes motion, so it must be pointing upwards if the ball is falling down.