Question

. A ball is thrown so that it just clears a 10 feet fence 60 feet away. If it left the hand 5 feet above the ground and at an angle of 60° with the horizontal, what was the initial velocity of the ball?

Answer 1

y = ax^2 + bx + c
y' = 2ax+b
y'(0) = tan60° = √3
so, b = √3

y = ax^2 + √3 x + c
y(0) = 5, so

y = ax^2 + √3 x + 5

y(60) = 10, so
3600a + 60√3 + 5 = 10
a = (1-12√3)/720

y = (1-12√3)/720 x^2 + √3 x + 5
= -0.027x^2 + 1.732x + 5

so,

g/(2v^2) sec^2 60° = 0.027
4.9 * 4 / 0.027 = v^2
v ≈ 27 m/s

Answer 2

. A ball is thrown so that it just clears a 10 feet fence 60 feet away. If it left the hand 5 feet above the ground and at an angle of 60° with the horizontal, what was the initial velocity of the ball?