Ball thrown up at speed 91feet/sec from height of 15 feet off the ground.The height h of ball after t seconds found with equation h=−16t^2+91t+15.When height be 102 feet?When will ball reach ground?

Question

An object is thrown upward at a speed of 91 feet per second by a machine from a height of 15 feet off the ground. The height h of the object after t seconds can be found using the equation h = − 16t^2 + 91t + 15 When will the height be 102 feet? When will the object reach the ground?

Answer 1

Answer 2

Plug 102 to the equation for the height of the object h with initial velocity = 91 and initial height = 15 we'll get:

102 = -16 t^2 + 91 t + 15 or -16 t^2 + 91 t - 87 = 0 and we get t = 1.216 or t =4.471

To Calculate when ball reach the ground, just plug zero to h and solve for t:

0 = -16 t^2 + 91 t + 15 therefore the answer for the time it reaches the ground is t = 5.847 (using quadratic Formula or the Graphing calculator)