## Calculus: Early Transcendentals 9th Edition

(a) After 2 seconds, the velocity of the projectile is $4.9~m/s$ After 4 seconds, the velocity of the projectile is $-14.7~m/s$ (b) The projectile reaches maximum height after $2.5$ seconds. (c) The maximum height is $32.6$ meters (d) The projectile hits the ground after $5.08$ seconds. (e) When the projectile hits the ground, the velocity is $-25.3~m/s$
(a) $h = 2+24.5t-4.9t^2$ $v = 24.5-9.8t$ After 2 seconds: $v = 24.5-9.8(2) = 4.9$ After 2 seconds, the velocity of the projectile is $4.9~m/s$ After 4 seconds: $v = 24.5-9.8(4) = -14.7$ After 4 seconds, the velocity of the projectile is $-14.7~m/s$ (b) When the projectile reaches the maximum height, the velocity is 0. $v = 24.5-9.8t = 0$ $t = \frac{24.5}{9.8}$ $t = 2.5$ The projectile reaches maximum height after $2.5$ seconds. (c) $h = 2+24.5t-4.9t^2$ $h = 2+24.5(2.5)-4.9(2.5)^2$ $h = 32.6$ The maximum height is $32.6$ meters (d) $h = 2+24.5t-4.9t^2 = 0$ $4.9t^2-24.5t-2 = 0$ We can use the quadratic formula to find the time $t$ when the projectile hits the ground: $t = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$ $t = \frac{-(-24.5) \pm \sqrt{(-24.5)^2-4(4.9)(-2)}}{(2)(4.9)}$ $t = \frac{24.5 \pm \sqrt{600.25+39.2}}{9.8}$ $t = -0.08~~$ or $~~t = 5.08$ The projectile hits the ground after $5.08$ seconds. (e) $v = 24.5-9.8t$ $v = 24.5-9.8(5.08) = -25.3$ When the projectile hits the ground, the velocity is $-25.3~m/s$