Question

Emma launches a ball off of a platform of height h with velocity v₀ at an angle θ above the horizontal. A wall with a ball-sized hole a hight of H above the floor, with H>h, is a horizontal distance L= 16.0 m away from Emma. The difference in height between H and h is one fifth of L, the square of the magnitude of the velocity, v₀^2, is five-fourths of the product gL.

Determine the two values of θ that ensure the ball passes through the hole.

Hint: It is possible to use the quadratic formula to solve any quadratic equation that is equal to zero, even when the variable itself is a function.

Answer 1

hole ginen L=16m V² - 9,8x 16 Vo = 12.5 m/s H: h= 16 & L=16m condition is that pays Condition is that ball passes through hole iff at position of hole vertical velocity of ball is o. Izero) vertical velocity of ball when projected is - Vosino wing newton's theird law Page No.

4²-42 = 2gs 0-12 sin20 = -2 X 9.84 16 = = R sin20= 2X 9.8 x 16 I displacement ish-b) 9.8 x 16 s sin?o= 2 = sino = + 1 2 3 when sino +5215 0= 39.23° when sino -1215 » Cos 17/+0) = 1445 => 0 =-39.236 so, corresponding values of o are £ 39.23° | But this is not physically feasible.