Question

A little confused on these Motion problems:

An airplane with a speed of 95.3 m/s is climbing upward at an angle of 38.1 ° with respect to the horizontal. When the plane's altitude is 542 m, the pilot releases a package. (a) Calculate the distance along the ground, measured from a point directly beneath the point of release, to where the package hits the earth. (b) Relative to the ground, determine the angle of the velocity vector of the package just before impact.

As an aid in working this problem, consult Interactive Solution 3.41. A soccer player kicks the ball toward a goal that is 25.0 m in front of him. The ball leaves his foot at a speed of 17.6 m/s and an angle of 32.0 ° above the ground. Find the speed of the ball when the goalie catches it in front of the net.

Chapter 03, Problem 15

A skateboarder shoots off a ramp with a velocity of 5.7 m/s, directed at an angle of 56° above the horizontal. The end of the ramp is 1.2 m above the ground. Let the x axis be parallel to the ground, the +y direction be vertically upward, and take as the origin the point on the ground directly below the top of the ramp. (a) How high above the ground is the highest point that the skateboarder reaches? (b) When the skateboarder reaches the highest point, how far is this point horizontally from the end of the ramp?

Answer 1

a)

Not confusing if you use equation of path

since, package comes down from 542m, y = -542 m. Putting rest of the things and solving for x

-542 = xtan38.1 - 4.9 x² / (95.3cos38.1)²
x = 1358m

b)

using law of conservation of energy

1/2mv^2 = 1/2mu^2 + mgh

1/2 v^2 = 1/2 u^2 + gh

= 1/2 x 95.3^2 + 9.8 x 542

v = 140.4 m/s

since ux = vx = 95.3cos38.1 = 75 m/s ....horizontal velocity doesnt change

angle = cos^-1(75/140.4) = 57.71 deg

I hope it helps in solving rest of the problems.