A basketball is thrown at 45° to the horizontal. The hoop is located 4.4 m away horizontally at a height of 1.1 m above the point of release. What is the required initial speed?

Answer 1

1.1 = Vi t - 4.9 t^2
t = 4.4/u
but u = Vi because tan 45 = 1

1.1 = u t - 4.9 t^2
1.1 = u t - 4.9 (4.4^2)/u^2

1.1 u^2 = u^2 (4.4) - 94.9

3.3 u^2 = 94.9
u = Vi = 5.36 m/s
sqrt(u^2+Vi^2) = 7.58 m/s