Thank u 5 Chapter 4 Problem 18 A tightly crumpled wad of pap

5. Chapter 4, Problem 18. A tightly crumpled wad of paper is tossed toward a wastebasket sitting against a wall. The paper is thrown toward the basket from a height of 1.25 m at an angle of 10.0 below the horizontal with a speed of 2.00m/s The wastebasket is 30.0cm tall, has a diameter of 25.0cm, and its center is located 75.0em from the paper\'s point of release. Where does the paper land? 6. Chapter 4, Problem 19. The end of a water slide is 2.00m above the surface of a pool. For the last several meters, the slide is inclined downward at 20.0°. A swimmer tucked in the cannonball position leaves the end of the slide at 3.50m/s a. How far horizontally from the end of the slide does the swimmer hit the water? b. How fast is the swimmer traveling just before hitting the water? 7. Chapter 4, Problem 26 A woman stands 15.0ft from a point directly below the basket when she shoots a free throw. As the ball leaves he hand, it is 6.00 ft above the ground traveling at an angle of 54.0°. the basket is loort above the floor, find the velocity she must give the ball to make a basket without hting the rim or the backboard. 8. Chapter 4, Problem 30. Take the moon\'s orbit around the earth as circular. a. Using data from Appendix 8 (page 1156), calculate the moon\'s centripetal acceleration in its orbit. b. Measuring from the earth\'s center, here on the earth\'s surface you are 1 earth-radius from earth. The moon is 60 times further away. Just for kicks, multiply your answer from part a by 602. Any thoughts?

Solution

initial speed v = 2.0 m/s

angle of inclination 10⁰

horizontal vel V_x = 2.0Cos(10) = 1.97 m/s

vertical vel v_y = 2.0Sin(10) = 0.35 m/s

motion of the projectile is given by

x= v_x t   = 1.97t

and

y = v_yt+gt²/2 = 0.35t+4.9t²

maximum horizontal distance = 75+12.5 = 87.5 cm to the wall

horizontal travel time =0. 875/1.97 = 0.44s

vertical descent = 0.35*0.44+4.9*(0.44)² = 1.12 m

By the time the ball reaches the wall it will descend 1.12 m i.e.

1.25-1.12 = 0.13 m above the ground inside the waste basket

The center is 0.75 m from the point of origin.

time of flight = 0.75/1.97 = 0.38 s

vertical descent in this time = 0.35*0.38+4.9*(0.38)² = 0.84 m

By the time the ball reached the center of the basket it is 1.25-0.84 = 0.41m above the ground where as the basket top edge is at 0.3 m from the ground.

The paper ball will land inside the basket

it is assumed the plane of projection is coincides with plane containing the vertical through the center of the basket and the vertical at the point of projection