Question

18) A ball is thrown across a playing field from a height of 5 ft above the ground at an angle of 45° to the horizontalat a speed of 20 ft/s. It can be deduced fromphysical principles thatthe path of the ball is modeled by the function f(x) = - 32 x +x+5 400 a. At what horizontal distance xis the ball 7 ft high? b. At what horizontal distance x does the ball hit the ground?

Answer 1

32 ☆ + x +5 400 a height eft f(x)= 7ft 7 - 32 x +x+5 400 32x +400x - 300=0 - b ± √5-yac 2a Here a = -32, b= 400 C= - 800

- 400+ √(400)? -4 (-32) (-800) 21-32) 400 + 577600 - 64 -400 240 -64 56 = -400 +240 and = -400-240 - 64 10 2 at x = 5 / 5 the ball is and x=10 Therefore, Eft high one value is ball moving upwards & other is when moving downwar.de when ball hits the ground, height - f(x)=0 -32 400 +*+5 = 0 - 32x + 4004+ 2000 =0 - b ± √br-yac 20 a = -32, Here b=uoo C= 2000 -400+ VILOOP-H(-32)(2000) 21-32) - 400 + 80/65 -64 y = 5(5+565) = and - 5(1655) 4 4 X= 16.32 -3.827 be It cannot negative when ball hits ground 16.32 ft Therefore, distance E 16 ft