Question

The height (in meters) of a projectile shot vertically upward from a point 2 m above ground level with an initial velocity of 22.5 m/s is h = 2 + 22.5t − 4.9t2 after t seconds. (Round your answers to two decimal places.)(a) Find the velocity after 2 s and after 4 s.v(2) = _____________m/sv(4) = ______________m/s(b) When does the projectile reach its maximum height?___________________s(c) What is the maximum height?________________________m(d) When does it hit the ground?______________________________s(e) With what velocity does it hit the ground?__________________________________m/s

Answer 1

a)
`v(2\,s) = 2.9\,(m)/(s)`, b)
`v(4\,s) = -16.7\,(m)/(s)`, c)
`t = 2.296\,s`, d)
`h (2.296\,s) = 27.829\,m`, e)
`t \approx 4.679\,s`, f)
`v(4.679\,s) = -23.354\,(m)/(s)`

Explanation:

The velocity function can be derived by the differentiating the height function:

`v = 22.5-9.8\cdot t`

Velocities after 2 and 4 seconds are, respectively:

a)
`v(2\,s) = 2.9\,(m)/(s)`

b)
`v(4\,s) = -16.7\,(m)/(s)`

The maximum height is reached when velocity is zero. Then:

`22.5-9.8\cdot t = 0`

c)
`t = 2.296\,s`

The maximum height is:

d)
`h (2.296\,s) = 27.829\,m`

The time required to hit the ground is:

`-4.9\cdot t^(2)+22.5\cdot t +2 = 0`

Roots of the second-order polynomial are:

`t_(1) \approx 4.679\,s`

`t_(2) \approx -0.087\,s`

Only the first root is physically reasonable.

e)
`t \approx 4.679\,s`

The velocity when the projectile hits the ground is:

f)
`v(4.679\,s) = -23.354\,(m)/(s)`