Question

A ball is launched into the air from a platform 14 feet off the ground with an initial velocity,v0, of 64 feet per second.The height H(t) of the ball at any time, t can be modeled by the formulae H(t)=-16t^2+v0t+h0 When will the ball hit the ground rounded to the nearest hundredth?At what time will the ball have a height of 62 feet during the descent?

Answer 1

a) After 4.21 seconds

b) t=1 or t=2 seconds

Explanation:

The height of the ball at anytime t, is modeled by:

`H(t)=-16t^2+v_0t+h_0` , where
`v_0=64` feet per second and
`h_0=14` feet.

We substitute the initial velocity and the initial height to obtain:

`H(t)=-16t^2+64t+14`

When the ball hit the ground, then the heigth becomes zero.

We equate the function to zero to obtain:

`-16t^2+64t+14=0`

We use the quadratic formula to obtain:

`t=(-b\pm√(b^2-4ac) )/(2a)`

where a=-16, b=64 and c=14.

We substitute to obtain:

`t=(-64\pm√((-64)^2-4*-16*14) )/(2*-16)`

This simplifies to:

`t=2\pm(78)/(4)`

`t=4.21` or
`t=-0.21`

We discard the negatively, value to get t=-4.21 seconds.

The ball will hit the ground after 4.21 seconds.

To find the time that the ball have a height of 62 feet during the descent, we set H(t)=62 and solve for t.

`-16t^2+64t+14=62\\-16t^2+64t+14-62=0`

`-16t^2+64t-48=0`

Divide through by 16 to get:

`t^2-4t+3=0`

`\implies (t-1)(t-3)=0`

`\implies t=1\:or\:t=3`

There the ball will first have a height of 62 at t=1 seconds and the second time is t=3 seconds