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

Answer:

6.21 seconds

Step-by-step 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,

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

$t=\frac{-35\pm \sqrt{35^2-4\times -16\times 400}}{2\times -16}$

$t=\frac{-35\pm \sqrt{1225+25600}}{-32}$

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

∵ Time cannot be negative,

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