On August 12, 1987, Alessandro Andrei set a world record in the shot put with a

Question

On August 12, 1987, Alessandro Andrei set a world record in the shot put with a throw of 22.9 m (75 ft 2 in.). What was the initial speed of the shot if he released it at a height of 2.10 m and threw it at an angle of 38 ? above the horizontal? (Although the maximum distance for a projectile on level ground is achieved at 45 ?, the ideal angle is smaller when the projectile lands at an altitude below its point of release; thus, 38 ? will give a longer range than 45 ? in the shot put.)

please show all work. I need help for understanding not an answer.

Explanation / Answer

Since this is such a classical question, I'm not going to actually solve all of it for you. But I will tell you *how* to solve it and leave the arithmetic to you. First, it's usually a good idea to draw a picture. Get a few pieces of graph paper and draw the origin in the lower left corner. Now go 'up' the y axis to the point that corresponds to 1.83 meters and make a dot. This dot is where the ball is released. Now go to 4.21 meters to the 'right' and up 3.05 meters and make another dot. This is the center of the basket. When the ball is released, it has an initial velocity of 4.88 m/s at an angle of 35 degrees above horizontal. Yhis means that its velocity can be broken down into two vextor components (vertical and horizontal or y and x) as follows: Vx = 4.88 cos(35) = 3.997 m/s horizontally Vy = 4.88 sin(35) = 2.799 m/s vertically Since there is no force acting on the ball horizontally, it will travel horizontally at a constant 3.997 m/s Vertically, it has an acceleration of -9.8 m/s/s (gravity) acting upon it. Since you're working this kind of problem, I assume that you've already had the equations of motion (involving position, velocity, acceleration, and time) along a straight line. This problem is no different than those problems except that you now have two lines along which you must calculate. The one (and only) variable that 'ties' the two axis together is time. So...... Remember that velocity under acceleration is given (s a function of time) by v(t) = v0 + at^2 If the vertical velocity is 2.799 m/s and gravity is acting 'downwards' (negative) at -9.8 m/s/s then the velocity of the ball along the vertical axis (as a function of time) is Vx(t) = 2.799 - 9.8 t^2 At the top of it's trajectory the balls intaneous velocity is zero and that happens at 0 = 2.799 - 9.8 t^2 or t = sqrt (2.799/9.8) = .534 seconds after the ball is released. How high is the ball? Remember that v^2 = v0^2 + 2ay and we're interested at v = 0 so 0 = 2.799^2 - 2*9.8*y so that y = (2.799^2)/(2*9.8) = .399 meters above the release point or .399 + 1.83 = 2.229 meters above the floor. In .534 seconds at a velocity of 3.997 m/s horizontally, it will travel .534 * 3.997 = 2.134 meters horizontally (to the right) This is the 'top' of the balls trajectory (2.134,2.229)

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