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

Answer:

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

Step-by-step 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) $v(2\,s) = 2.9\,\frac{m}{s}$

b) $v(4\,s) = -16.7\,\frac{m}{s}$

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) $v(4.679\,s) = -23.354\,\frac{m}{s}$