1. A hot-air balloon is rising upward with a constant speed of 2.50 m/s. When th

Question

1. A hot-air balloon is rising upward with a constant speed of 2.50 m/s. When the balloon is 20.0 m above the ground, the balloonist accidentally drops a compass over the side of the balloon. Find the time it takes for the compass to hit the ground. In a diagram, show the path taken by the compass, as seen by a person on the ground. In the same diagram, choose five different points on the path of the compass and draw the velocity vector and the acceleration vector of the compass at each of the five points.

2. A 20.0 kg sled is being pulled across the horizontal surface of a frozen lake at a constant velocity of 0.50 m/s. The pulling force has a magnitude of 80.0 N and is directed at an angle of 300 above the horizontal. Find the coefficient of kinetic friction. Note: Be sure to first draw a diagram showing the forces on the sled, and then draw a free-body diagram.

3. A soccer player kicks a ball toward a goal that is 16.8 m in front of him. The ball leaves his foot at a speed of 16.0 m/s and an angle 28.00 above horizontal. Fine the speed of the ball when the goalie catches it in front of the net. Assuming the ball was on the ground when the player kicked it, how high above the ground did the goalie catch the ball? In a diagram, show the path taken by the ball, then choose five different points on the path of the ball and draw the velocity vector and the acceleration vector of the ball at each of the five points.


The diagrams are vital if you want the points. Show work clearly, and you will be rewarded generously.

Explanation / Answer

1.

Solve this quadratic equation for t:

y = 2.50t -4.9t^2 +20 = 0

Take the positive one of the two roots.

t=2.291 sec

first it will go upside sometime then straight down to ground basically in same line as there is no acceleration

velocity vector upside or downside and acceleration will be to the ground all time.

2.

forces in horizontal direction are : 80cos(30)-frictional force
as it is pulled with constant velocity, acceleration =0

so, no net force acts in x-direction

so, frictional force = 80cos(30) = %u03BCmg

put m= 20 and g = 9.8

so, co-eff of kinetic friction %u03BC = 80cos(30)/20*9.8 = 0.10204

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