Question

The position function for a free-falling object is s(t) = -16t^2 + vt +5. A ball is thrown upward from the top of a 400-foot building with an initial velocity of 35 feet per second. At what time will the ball reach the ground (round answer to nearest hundredths)? O at 2.19 seconds at 0 seconds at 6.21 seconds at -4.02 seconds

Answer 1

6.21 seconds

Explanation:

The position function of a free-falling object is,

`s(t) = -16t^2+v_0t+s_0`

Where,

`s_0` = initial height of the object,

`v_0` = initial velocity of the object,

Here,

`s_0=400\text{ feet}`

`v_0=35\text{ feet per sec}`

Hence, the height of the ball after t seconds,

`s(t)=-16t^2+35t+400`

When s(t) = 0,

`\implies -16t^2+35t+400=0`

`t=(-35\pm √(35^2-4* -16* 400))/(2* -16)`

`t=(-35\pm √(1225+25600))/(-32)`

`\implies t\approx -4.02\text{ or }t\approx 6.21`

∵ Time cannot be negative,

Hence, after 6.21 seconds the ball will reach the ground.