A major leaguer hits a baseball so that it leaves the bat at a speed of 28.0m/s

Question

A major leaguer hits a baseball so that it leaves the bat at a speed of 28.0m/s and at an angle of 36.1 above the horizontal. You can ignore air resistance.

PartA) At what two times is the baseball at a height of 8.50 above the point at which it left the bat?
PartB) Calculate the horizontal component of the baseball's velocity at each of the two times you found in part (a).
PartC) Calculate the vertical component of the baseball's velocity at each of the two times you found in part (a).
PartD) What is the magnitude of the baseball's velocity when it returns to the level at which it left the bat?
PartE) What is the direction of the baseball's velocity when it returns to the level at which it left the bat?

Please explain how to solve... thanks :)

Explanation / Answer

the initial velocity of ball, u = 28m/s Angle of projection, theta = 36.9 let y be the vertical height of the ball at any stage and let x be its corresponding horizontal distance By parabolic path equation, y = x tan(theta) - {g/2u^2cos^2(theta)} x^2 => 8.50 = x 0.75 - [9.8/2 x 28x 28 x 0.624] x^2 => 8.5 = 0.75x - [9.8/978.4320]x^2 => 8.5 = 0.75x - 0.01x^2 Solving the equation for x we get, x = 13.91m Now, horizontal distance = 13.91m horizontal component of velocity = ucos(theta) = 28 x 0.79 = 22.12m/s a = 0, there is no horizontal component for g By newton;s laws of motion, distance, s = ut + 1/2at^2 =>13.91 = 22.12t => t = 13.91/22.12 => t = 0.62s....(ANSWER) We know that during the two times at which the ball will be at the same height., in the first time the ball will move upward while in the second the ball moves downwards, hence the horizontal distance at which the ball is vertically high at 8.5m while moving downward will be 13.91m from the point of landing....but we need the horizontal distance from the point of projection, therefore, the total range of ball minus the distance from the point of landing will give its distance from the point of projection Range = u^2sin2(theta)/g = 28 x 28 x 0.96/9.8 = 76.8 Horizontal distance of ball in downward motion from point of projection = range - horizontal distance from point of landing => x = 76.8 - 13.91 => x = 62.89 By newton's laws of motion, S = ut + 1/2at^2 a = 0 => s = ut => t = s/u => t = 62.89/22.12 => t = 2.8s.........(ANSWER) sorrry friend ,,only one question to be answered at a time!

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