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Oil SPS 2 1.pdf THE WALL (10 points total) You throw a ball toward a wall at speed vo 25.0 m/s and at angle θ = 40.0° above the horizontal (figure bclow). The wall is distance d - 22.0 m from the rclcasc point of the ball Assumc no air rcsistancc 1. How far above the release point h does the ball hit the wall? Start your argument from the basic equations of motion into the horizontal and vertical directions. (2 pts) as the x axis and the release point to be the origin. (2 pts) y axis and the release point to be the origin. (2 pts) 2. What is the horizontal component of its velocity as the ball hits the wall? Set the horizontal 3. What is the vertical component of its velocity as the ball hits the wall? Set the vertical as the 4. When the ball hits the wall, has it passed the highest point on its trajectory? (2 pts) 5. Assuming thc ball bounccs off the wall such that it now has thc cxact samc spccd as the instant it hit the wall, but with the horizontal component of the velocity reversed, how far away from the wall does the ball pass y-0? (2 pts) HALLIDAY, REsNiCK, And WalKer (HRW) O Type here to search 125 AM 1/25/2019 癖門: LNG

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Answer #1

a) we know the formula

y=voy*t+1/2at2............1)

t=d/vox=22/19.15=1.15s

from the figure

vox=vo*cos40=25*cos40=19.15m/s

voy=vo*sin40=25*sin40=16.07m/s

so putting the values the in equation 1

y=16.07*1.15+1/2*9.8*1.152=12m

so the answer is 12 m

b) we have already found out vox

vox=vo*cos40=25*cos40=19.15m/s

c) we have

voy=vo*sin40=25*sin40=16.07m/s

d) no it has not reached it highest point because the vertical component is still in positive so it hits the wall before reaching the highest point.

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