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Projectile, Range, Maximum Height, Time of Flight Problem 
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#1
Sep2108, 11:15 PM

P: 1

1. The problem statement, all variables and given/known data
Your teacher tosses a basketball. The ball gets through the hoop. How long does it take the ball to reach its maximum height? How long does it take the ball to reach the hoop? What is the horizontal length of the shot? (Neglect air friction). Initial velocity: 17 m/s 53 degree angle between the ball's initial position and the ground. The height of the teacher is 2.576 m. (This is the ball's initial height.) The height of the goal is 3.048 m. (This is the ball's final height.) 2. Relevant equations Vf = Vyo + at Vyo = VoSinΘ 3. The attempt at a solution I solved the first part as follows: Vyo = VoSinΘ Vf = Vyo + at 0 = 13.5768  9.8t 13.5768 = 9.8t t = 1.3854s (Time to get to maximum height. This answer is correct.) I'm having some trouble with the second and third part though. For the second part: Vyo = VoSinΘ d = Vyot + 1/2at^2 I subtracted the two heights (3.048m  2.576m = 0.472m). 0.472m = 13.5768t + 1/2(9.8m/s^2)t^2 I then used the quadratic formula: 13.5768 + (square root of: ((13.5768^2)  4(4.9)(0.472))/(2)/(0.472) The answer I derived was not correct (28.4). For the third part: Xmax = 2Vo^2sinΘcosΘ/g 2(17^2)(sin53)(cos53)/9.8 The answer I derived was not correct (28.3). Please help me if you can! 


#2
Sep2208, 09:03 PM

P: 159

For the second part, first calculate the time it takes the ball to reach maximum height, which you already found to be 1.4 seconds.
Next, calculate the maximum height it reached. From there, the object is falling the maximum height PLUS the .472 meter difference. Use that total height to calculate the total time. Because the object is simply falling from the maximum height, Vo is 0 and so it's not a quadratic equation. Once you find this time, add that to the first time you got. This will give you the total time. Once you have the total time, you can simply use S = VT to solve for the horizontal distance. Hope that helps :) 


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