the actual question
A 3 point jump shot is released 2.2 m above the ground, 5.97 m from the basket, which is 2.99 m high. For launch angles of 30° and 60°, find the speed needed to make the basket.
2007-01-03 13:35:55 · 1 answers · asked by Cameron F 1 in Science & Mathematics ➔ Physics
The shot must travel a vertical distance of 2.99 - 2.2 = .79m. It must travel a horizontal distance of 5.97m For a given launch velocity of V and angle ø, the time "t" it takes to make the horizontal travel is 5.97/(V*cosø). (The horizontal velocity is V*cosø.) In that time it travels a vertical distance of V*t*sinø - .5*g*t^2. Solve for t first, then put that into the equation .79 = V*t*sinø - .5*g*t^2 and solve for V for the values of ø given:
.79 = V*5.97*sinø/(V*cosø) - .5*g*[5.97/(V*cosø)]^2
.79 = 5.97*tanø - .5*5.97^2*g/[V*cosø]^2
To see the solution for the general case and the numerical results, go to http://img245.imageshack.us/img245/513/balllaunchanglepi1.png
2007-01-03 14:35:42 · answer #1 · answered by gp4rts 7 · 0⤊ 0⤋