Having a hard time solving last part

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The discussion focuses on the physics problem of a ball thrown at a speed of 24.0 m/s and an angle of 43.0° towards a wall located 16.0 m away. The ball hits the wall at a height of 10.7 m above the release point, with a horizontal velocity component of 17.5 m/s. The vertical velocity component at impact requires calculating the effect of gravity on the initial vertical velocity, which is determined using the sine function of the launch angle. The correct approach involves vector addition of the horizontal and vertical components to find the resultant velocity.

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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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