Question

The height (in meters) of a projectile shot vertically upward from a point 2 m above ground level with an initial velocity of 25.5 m/s is h = 2 + 25.5t ? 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) = 5.9 Correct: Your answer is correct. m/s v(4) = -13.7 Correct: Your answer is correct. m/s (b) When does the projectile reach its maximum height? 2.60 Correct: Your answer is correct. s (c) What is the maximum height? 37.2 Incorrect: Your answer is incorrect. m (d) When does it hit the ground? 5.36 Incorrect: Your answer is incorrect. s (e) With what velocity does it hit the ground? 27.03 Incorrect: Your answer is incorrect. m/s

Answer 1

Answer:

a) $v(2\,s) = 5.9\,\frac{m}{s}$, $v(4\,s) = -13.7\,\frac{m}{s}$, b) $t = 2.602\,s$, c) $h(2.602\,s) = 35.176\,m$, d) $t = 5.281\,s$, e) $v(2.602\,s) = -26.254\,\frac{m}{s}$

Explanation:

a) The velocity function is determined by deriving the position function in time:

$v(t) = 25.5-9.8\cdot t$

Velocities after 2 seconds and 4 seconds are, respectively:

$v(2\,s) = 5.9\,\frac{m}{s}$

$v(4\,s) = -13.7\,\frac{m}{s}$

b) The maximum height is reached when velocity is equal to zero:

$25.5-9.8\cdot t = 0$

The time when the projectile reaches the maximum height:

$t = 2.602\,s$

c) The maximum height is:

$h (2.602\,s) = 2 + 25.5\cdot (2.602\,s)-4.9\cdot (2.602\,s)^{2}$

$h(2.602\,s) = 35.176\,m$

d) The projectile hits the ground when height is equal to zero:

$-4.9\cdot t^{2}+25.5\cdot t + 2 =0$

The roots of the second order polynomial are presented below:

$t_{1} \approx 5.281\,s$

$t_{2} \approx -0.077\,s$

The first one is the only reasonable solution in physical terms.

$t = 5.281\,s$

e) The velocity of the projectile when it hits the ground is:

$v(2.602\,s) = 25.5-9.8\cdot (5.281\,s)$

$v(2.602\,s) = -26.254\,\frac{m}{s}$