Isaac throws an apple straight up (in the positive direction) from 1.0 m above the ground, reaching a maximum height of 35 meters. Neglecting air resistance, what is the ball’s velocity when it hits the ground?

Answer 1

h(t) = 1 + vt - 16t^2

max height is reached at t = v/32, so

1 + v(v/32) - 16(v/32)^2 = 35
v = 8√34

h(t) = 1 + 8√34 t - 16t^2
h=0 at t=2.94

v(t) = 8√34 - 32t, so when h=0,
v = 8√34 - 32(2.94) = -47.43 ft/s

Answer 2

V^2=Vo^2+2a(Xf-Xo)
=O^2+2(-9.8)(35-0)
Vf^2=-686
Square root 686 which equals
Vf= -26.2 m/s

Answer 3

V^2=Vo^2+2a(Xf-Xo)
=O^2+2(-9.8)(35-0)
Vf^2=-686
Square root 686 which equals
Vf= -26.2 m/s

in this equation, why is it 35-0 not 35-1?? if its thrown initially 1m from the ground