Having a hard time solving last part

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The discussion focuses on a physics problem involving a ball thrown at a speed of 24.0 m/s and an angle of 43.0° toward a wall 16.0 m away. The correct height at which the ball hits the wall is determined to be 10.7 m above the release point. The horizontal component of the ball's velocity upon impact is calculated as 17.5 m/s. Participants emphasize the need to account for gravity's effect on the vertical velocity component at the moment of impact. The final vertical velocity component requires adjusting the initial vertical component due to gravitational deceleration.
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You throw a ball toward a wall at speed 24.0 m/s and at angle θ0 = 43.0° above the horizontal (Fig. 4-35). The wall is distance d = 16.0 m from the release point of the ball. (a) How far above the release point does the ball hit the wall? What are the (b) horizontal and (c) vertical components of its velocity as it hits the wall?


(a)Number 10.7 Units m correct
(b)Number 17.5 Units m/s correct ---- cos(43)* 24
(c)Number Units m/s -- can not get --- attempt-- sin(43) *24
 
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mikenash said:
You throw a ball toward a wall at speed 24.0 m/s and at angle θ0 = 43.0° above the horizontal (Fig. 4-35). The wall is distance d = 16.0 m from the release point of the ball. (a) How far above the release point does the ball hit the wall? What are the (b) horizontal and (c) vertical components of its velocity as it hits the wall?


(a)Number 10.7 Units m correct
(b)Number 17.5 Units m/s correct ---- cos(43)* 24
(c)Number Units m/s -- can not get --- attempt-- sin(43) *24

What is your vertical velocity component when it strikes the wall? Remember at the time it strikes the wall it has been slowed by gravity and will no longer be the Initial vertical component. When you figure that, you can add the component vectors to arrive at the result.
 
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