Question

A projectile is launched from ground level with an initial velocity of v 0 feet per second. Neglecting air? resistance, its height in feet t seconds after launch is given by s equals negative 16 t squared plus v 0 t. Find the? time(s) that the projectile will? (a) reach a height of 80 ft and? (b) return to the ground when v 0 is 32 feet per second. ?(a) Find the? time(s) that the projectile will reach a height of 80 ft when v 0equals32 feet per second. Select the correct choice below? and, if? necessary, fill in the answer box to complete your choice.

Answer 1

S = - 16 t^2 + V_o t V_o = 32 ft/s therefore s = - 16 t^2 + 32 t a) To reach height of 80 ft therefore - 16 t^2 + 32t = 80 16t^2 - 32t + 80 = 0 t^2 - 2t + 5 = 0 t = - (- 2) pm squareroot (- 2) ^2 - 4(1) (5) /2(1) = 2 pm squareroot 4 - 20/2 2 pm squareroot - 16/2 therefore time is imaginary, it will never reach height of 80 ft or maximum height in at t = - b/2a = - 32/2(- 16) t = 1 S(1) = - 16(1) ^2 + 32(1) = 16 ft rightarrow maximum height \r\n b) Return to ground S = 0 therefore - 16 t^2 + 32t = 0 - 16t(t + 2) = 0 t = 2 or t = 0 therefore After 2 seconds