a child throws a rock with a velocity of 25 m/s at angle of 65 degrees above the horizontal. the rock hits the side of the building at a height of 12 meters. (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? (d) When it hits, has it passed the highest point on its trajectory?

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qwterabite qwterabite · Answer 1 · 2022-12-19T18:31:11+01:00

(a) The ball hits the wall 5.3 meters above the release point.

(b) The horizontal component of its velocity is 17.5 m/s. (c) The vertical component of its velocity is 17.1 m/s.

(d) Yes, it has passed the highest point on its trajectory.

(a) Using the equation for range, we can solve for time:

R = V*t*cosθ =>25*t*cos65 = 12 t = 0.75 s

We can then solve for the height using the equation for vertical displacement:

h = V*t*sinθ - 0.5*g*t^2

h = 25*0.75*sin65 - 0.5*9.8*0.75^2 h

= 5.3 m

(b) We can solve for the horizontal component of the velocity using the equation for range:

R = V*t*cosθ 17.5

= 25*t*cos65 t

= 0.7 s V = 17.5 m/s

(c) We can solve for the vertical component of the velocity using the equation for vertical displacement:

h = V*t*sinθ - 0.5*g*t^2 17.1

= 25*0.75*sin65 - 0.5*9.8*0.75^2 V = 17.1 m/s

(d) Yes, the ball has passed the highest point on its trajectory because the ball is traveling downwards when it hits the wall.

To know more about trajectory refer to the link brainly.com/question/28874076

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