Question

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.

Answer 1

Answer:

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.