Two stones are thrown vertically upward with matching initial velocities of 48ft/s at timet = 0. One stone is thrown from the edge of a bridge that is 32 ftabove the ground and the other stone is thrown from ground level. The heightof the stone thrown from the bridge aftertseconds isf (t) = -16t2 + 48t + 32, and the height of the stone thrown from the ground aftertseconds isg (t) = -16t2 + 48t.

Step-by-step explanation:

Stone thrown from ground:

S = - 16t² + 48t + 32

Velocity = V = dS/dt = - 32t + 48

Acceleration = dV/dr = - 32

Because Velocity is 0 at the top, we have V = 0

That is

-32t + 48 = 0

t = 48/32 = 3/2

S = - 16 (3/2) ² + 48 (3/2) + 32

= 36 + 72 + 32

S = 140 ft ... (1)

Stone thrown above the bridge level:

S = - 16t² + 48t

Velocity = V = dS/dt = - 32t + 48

Acceleration = dV/dr = - 32

Because Velocity is 0 at the top, we have V = 0

That is

-32t + 48 = 0

t = 48/32 = 3/2

S = - 16 (3/2) ² + 48 (3/2)

= 36 + 72

S = 108ft ... (2)

From (1) and (2)

140 - 108 = 32ft

Which shows that the stone then from the ground levels is 32ft higher than the one thrown at the edge of the bridge.